Given $lim_{x rightarrow a}f(x)+frac1{left|f(x)right|}=0$, what is $lim_{x rightarrow a}f(x)$?












4












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Another question I am struggling with:



Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




Thank you all for your answers!!



But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






calculus limits







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$endgroup$

















    4












    $begingroup$


    Another question I am struggling with:



    Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




    Thank you all for your answers!!



    But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






    calculus limits







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Another question I am struggling with:



      Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




      Thank you all for your answers!!



      But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






      calculus limits







      share|cite|improve this question











      $endgroup$




      Another question I am struggling with:



      Let $f:mathbb{R} rightarrow mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $lim_{x rightarrow a}left(f(x)+dfrac{1}{left|f(x)right|}right)=0$. Find $lim_{x rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.




      Thank you all for your answers!!



      But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there?






      calculus limits




      calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 13:30









      Did

      248k23225464




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      asked Nov 30 '13 at 11:27









      user83081user83081

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          2 Answers
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          6












          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47



















          4












          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47











          Your Answer





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          6












          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          6












          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47














          6












          6








          6





          $begingroup$

          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.






          share|cite|improve this answer











          $endgroup$



          If $f(x)$ tends to a limit as $x to a$ then this limit must be $-1$. First we note that the limit $lim_{x to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $infty$. Now if we have the limit as $L < 0$, then $|f(x)| to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '13 at 12:02

























          answered Nov 30 '13 at 11:38









          Paramanand SinghParamanand Singh

          50.8k557168




          50.8k557168












          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47


















          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47




          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47











          4












          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          4












          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47














          4












          4








          4





          $begingroup$

          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$






          share|cite|improve this answer











          $endgroup$



          Graphing the auxiliary function $$phi:quad umapsto u+{1over |u|}qquad(uindot{mathbb R})$$
          we see that (a) $ |phi(u)|geq{5over6}$ when $uleq-{3over2}$ or $ugeq-{1over2}$, and that (b) the function $phi$ is continuous and strictly increasing on the interval $I: -{3over2}<u<-{1over2}$. From (b) we conclude that $phirestriction I$ has an inverse $phi^{-1}$ which is continuous on $phi(I)$.



          Now we are told that the function
          $$g(x):=f(x)+{1over |f(x)|}=phibigl(f(x)bigr)$$
          has limit $0$ when $xto a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5over6}$ for all $xindot U$. According to (a) above this enforces $-{3over2}<f(x)<-{1over2}$ for all $xindot U$. It follows that $f(x)=phi^{-1}bigl(g(x)bigr)$ for all $xindot U$, so that we obtain
          $$lim_{xto a}f(x)=lim_{xto a}left(phi^{-1}bigl(g(x)bigr)right)=phi^{-1}(0)=-1 .$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '13 at 13:40

























          answered Nov 30 '13 at 12:46









          Christian BlatterChristian Blatter

          175k8115327




          175k8115327












          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47


















          • $begingroup$
            Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
            $endgroup$
            – user83081
            Nov 30 '13 at 15:47
















          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47




          $begingroup$
          Thank you for your answer! But.. How can I prove by definition that this is indeed the limit? take $epsilon>0$, I have to show that there exists a $delta>0$ such that, $|x-a| leq delta$ implies $|f(x)+1| leq epsilon$. now, I do know that there exists an $delta_1>0$, such that $|f(x)+frac{1}{|f(x)|}|< epsilon$.Or, $frac{epsilon}{k}$ for some constant $k$ But, how can I continue from there? Thanks!
          $endgroup$
          – user83081
          Nov 30 '13 at 15:47


















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          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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          File:DeusFollowingSea.jpg

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          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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