How can I evaluate the left limit of this function?












1












$begingroup$


If $f(x) = {{sqrt[4]{x^2tan^2 4x}}over{2x}}$



How can I evaluate
$$lim_{xto 0^-} f(x)$$



Here's what I tried:



begin{align}
lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{2x}}
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{x}} \
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{-sqrt[4]{x^4}}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}}
end{align}



In the second step I wrote $x$ as $-sqrt[4]{x^4}$ because we are working with the left limit which is on the left of $0$



begin{align}
-{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}} & = -{1over 2}lim_{xto 0^-} sqrt[4]{{tan^2 4x}over{x^2}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{left({tan 4x}over{x}right)^2} \
& = -{1over 2} sqrt[4]{left( lim_{xto 0^-} {{tan 4x}over{x}}right)^2} \
& = -{1over 2} sqrt[4]{(4)^2} \
& = -{1over 2} sqrt[4]{16} \
& = -{1over 2} × 2 = -1
end{align}



Are all these steps correct?
I feel something wrong about the second step, because I found some other people who solved it $1$ and not $-1$ as my solution brought me.






real-analysis calculus limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    No I haven't but if you look at the tags, you will see that I don't want to use l'Hopital
    $endgroup$
    – Yousef Essam
    Dec 29 '18 at 20:06








  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – user458276
    Dec 29 '18 at 20:08










  • $begingroup$
    In that case, you'll probably need to split up tan into sin/cos, and use the usual identities for $lim_{xto 0}sin(x)/x=1$ and $lim_{xto 0}(1-cos(x))/x=0$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 20:09










  • $begingroup$
    Take the $2x$ inside the root and cancel the $x^2$. Denominator will become $16x^2$. Then use $lim_{xto 0}{frac{tan(x)}{x}}=1$
    $endgroup$
    – Sauhard Sharma
    Dec 29 '18 at 20:10










  • $begingroup$
    Only that with sine @Ben.
    $endgroup$
    – user376343
    Dec 29 '18 at 20:11
















1












$begingroup$


If $f(x) = {{sqrt[4]{x^2tan^2 4x}}over{2x}}$



How can I evaluate
$$lim_{xto 0^-} f(x)$$



Here's what I tried:



begin{align}
lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{2x}}
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{x}} \
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{-sqrt[4]{x^4}}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}}
end{align}



In the second step I wrote $x$ as $-sqrt[4]{x^4}$ because we are working with the left limit which is on the left of $0$



begin{align}
-{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}} & = -{1over 2}lim_{xto 0^-} sqrt[4]{{tan^2 4x}over{x^2}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{left({tan 4x}over{x}right)^2} \
& = -{1over 2} sqrt[4]{left( lim_{xto 0^-} {{tan 4x}over{x}}right)^2} \
& = -{1over 2} sqrt[4]{(4)^2} \
& = -{1over 2} sqrt[4]{16} \
& = -{1over 2} × 2 = -1
end{align}



Are all these steps correct?
I feel something wrong about the second step, because I found some other people who solved it $1$ and not $-1$ as my solution brought me.






real-analysis calculus limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    No I haven't but if you look at the tags, you will see that I don't want to use l'Hopital
    $endgroup$
    – Yousef Essam
    Dec 29 '18 at 20:06








  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – user458276
    Dec 29 '18 at 20:08










  • $begingroup$
    In that case, you'll probably need to split up tan into sin/cos, and use the usual identities for $lim_{xto 0}sin(x)/x=1$ and $lim_{xto 0}(1-cos(x))/x=0$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 20:09










  • $begingroup$
    Take the $2x$ inside the root and cancel the $x^2$. Denominator will become $16x^2$. Then use $lim_{xto 0}{frac{tan(x)}{x}}=1$
    $endgroup$
    – Sauhard Sharma
    Dec 29 '18 at 20:10










  • $begingroup$
    Only that with sine @Ben.
    $endgroup$
    – user376343
    Dec 29 '18 at 20:11














1












1








1





$begingroup$


If $f(x) = {{sqrt[4]{x^2tan^2 4x}}over{2x}}$



How can I evaluate
$$lim_{xto 0^-} f(x)$$



Here's what I tried:



begin{align}
lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{2x}}
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{x}} \
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{-sqrt[4]{x^4}}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}}
end{align}



In the second step I wrote $x$ as $-sqrt[4]{x^4}$ because we are working with the left limit which is on the left of $0$



begin{align}
-{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}} & = -{1over 2}lim_{xto 0^-} sqrt[4]{{tan^2 4x}over{x^2}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{left({tan 4x}over{x}right)^2} \
& = -{1over 2} sqrt[4]{left( lim_{xto 0^-} {{tan 4x}over{x}}right)^2} \
& = -{1over 2} sqrt[4]{(4)^2} \
& = -{1over 2} sqrt[4]{16} \
& = -{1over 2} × 2 = -1
end{align}



Are all these steps correct?
I feel something wrong about the second step, because I found some other people who solved it $1$ and not $-1$ as my solution brought me.






real-analysis calculus limits-without-lhopital







share|cite|improve this question











$endgroup$




If $f(x) = {{sqrt[4]{x^2tan^2 4x}}over{2x}}$



How can I evaluate
$$lim_{xto 0^-} f(x)$$



Here's what I tried:



begin{align}
lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{2x}}
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{x}} \
& = {1over 2}lim_{xto 0^-} {{sqrt[4]{x^2tan^2 4x}}over{-sqrt[4]{x^4}}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}}
end{align}



In the second step I wrote $x$ as $-sqrt[4]{x^4}$ because we are working with the left limit which is on the left of $0$



begin{align}
-{1over 2}lim_{xto 0^-} sqrt[4]{{x^2tan^2 4x}over{x^4}} & = -{1over 2}lim_{xto 0^-} sqrt[4]{{tan^2 4x}over{x^2}} \
& = -{1over 2}lim_{xto 0^-} sqrt[4]{left({tan 4x}over{x}right)^2} \
& = -{1over 2} sqrt[4]{left( lim_{xto 0^-} {{tan 4x}over{x}}right)^2} \
& = -{1over 2} sqrt[4]{(4)^2} \
& = -{1over 2} sqrt[4]{16} \
& = -{1over 2} × 2 = -1
end{align}



Are all these steps correct?
I feel something wrong about the second step, because I found some other people who solved it $1$ and not $-1$ as my solution brought me.






real-analysis calculus limits-without-lhopital




real-analysis calculus limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 20:32







Yousef Essam

















asked Dec 29 '18 at 20:00









Yousef EssamYousef Essam

335




335












  • $begingroup$
    No I haven't but if you look at the tags, you will see that I don't want to use l'Hopital
    $endgroup$
    – Yousef Essam
    Dec 29 '18 at 20:06








  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – user458276
    Dec 29 '18 at 20:08










  • $begingroup$
    In that case, you'll probably need to split up tan into sin/cos, and use the usual identities for $lim_{xto 0}sin(x)/x=1$ and $lim_{xto 0}(1-cos(x))/x=0$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 20:09










  • $begingroup$
    Take the $2x$ inside the root and cancel the $x^2$. Denominator will become $16x^2$. Then use $lim_{xto 0}{frac{tan(x)}{x}}=1$
    $endgroup$
    – Sauhard Sharma
    Dec 29 '18 at 20:10










  • $begingroup$
    Only that with sine @Ben.
    $endgroup$
    – user376343
    Dec 29 '18 at 20:11


















  • $begingroup$
    No I haven't but if you look at the tags, you will see that I don't want to use l'Hopital
    $endgroup$
    – Yousef Essam
    Dec 29 '18 at 20:06








  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – user458276
    Dec 29 '18 at 20:08










  • $begingroup$
    In that case, you'll probably need to split up tan into sin/cos, and use the usual identities for $lim_{xto 0}sin(x)/x=1$ and $lim_{xto 0}(1-cos(x))/x=0$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 20:09










  • $begingroup$
    Take the $2x$ inside the root and cancel the $x^2$. Denominator will become $16x^2$. Then use $lim_{xto 0}{frac{tan(x)}{x}}=1$
    $endgroup$
    – Sauhard Sharma
    Dec 29 '18 at 20:10










  • $begingroup$
    Only that with sine @Ben.
    $endgroup$
    – user376343
    Dec 29 '18 at 20:11
















$begingroup$
No I haven't but if you look at the tags, you will see that I don't want to use l'Hopital
$endgroup$
– Yousef Essam
Dec 29 '18 at 20:06






$begingroup$
No I haven't but if you look at the tags, you will see that I don't want to use l'Hopital
$endgroup$
– Yousef Essam
Dec 29 '18 at 20:06






1




1




$begingroup$
What have you tried so far?
$endgroup$
– user458276
Dec 29 '18 at 20:08




$begingroup$
What have you tried so far?
$endgroup$
– user458276
Dec 29 '18 at 20:08












$begingroup$
In that case, you'll probably need to split up tan into sin/cos, and use the usual identities for $lim_{xto 0}sin(x)/x=1$ and $lim_{xto 0}(1-cos(x))/x=0$.
$endgroup$
– Ben W
Dec 29 '18 at 20:09




$begingroup$
In that case, you'll probably need to split up tan into sin/cos, and use the usual identities for $lim_{xto 0}sin(x)/x=1$ and $lim_{xto 0}(1-cos(x))/x=0$.
$endgroup$
– Ben W
Dec 29 '18 at 20:09












$begingroup$
Take the $2x$ inside the root and cancel the $x^2$. Denominator will become $16x^2$. Then use $lim_{xto 0}{frac{tan(x)}{x}}=1$
$endgroup$
– Sauhard Sharma
Dec 29 '18 at 20:10




$begingroup$
Take the $2x$ inside the root and cancel the $x^2$. Denominator will become $16x^2$. Then use $lim_{xto 0}{frac{tan(x)}{x}}=1$
$endgroup$
– Sauhard Sharma
Dec 29 '18 at 20:10












$begingroup$
Only that with sine @Ben.
$endgroup$
– user376343
Dec 29 '18 at 20:11




$begingroup$
Only that with sine @Ben.
$endgroup$
– user376343
Dec 29 '18 at 20:11










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$begingroup$

I would write $$lim_{xto 0^-}frac{sqrt{|xtan(4x)|}}{2x}$$






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    $begingroup$

    I would write $$lim_{xto 0^-}frac{sqrt{|xtan(4x)|}}{2x}$$






    share|cite|improve this answer









    $endgroup$


















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      $begingroup$

      I would write $$lim_{xto 0^-}frac{sqrt{|xtan(4x)|}}{2x}$$






      share|cite|improve this answer









      $endgroup$
















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        $begingroup$

        I would write $$lim_{xto 0^-}frac{sqrt{|xtan(4x)|}}{2x}$$






        share|cite|improve this answer









        $endgroup$



        I would write $$lim_{xto 0^-}frac{sqrt{|xtan(4x)|}}{2x}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 20:12









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        73.5k42864




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