Limit expression using approximations
$begingroup$
I have 2 questions that I'm unsure how to answer, they are both related to same question and function.
I am given a function:
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.
Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.
Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.
Thank you so much for your help in advance.
calculus limits approximation
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
$endgroup$
add a comment |
$begingroup$
I have 2 questions that I'm unsure how to answer, they are both related to same question and function.
I am given a function:
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.
Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.
Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.
Thank you so much for your help in advance.
calculus limits approximation
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
$endgroup$
add a comment |
$begingroup$
I have 2 questions that I'm unsure how to answer, they are both related to same question and function.
I am given a function:
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.
Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.
Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.
Thank you so much for your help in advance.
calculus limits approximation
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
$endgroup$
I have 2 questions that I'm unsure how to answer, they are both related to same question and function.
I am given a function:
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.
Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.
Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.
Thank you so much for your help in advance.
calculus limits approximation
calculus limits approximation
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
edited Jan 6 at 6:10
Paramanand Singh
50.1k556163
50.1k556163
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
asked Jan 5 at 23:38
The StatisticianThe Statistician
96111
96111
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
We have that
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
and so $f(x)approx f(0)+xf'(0).$ Since
$$f'(0) = frac{e-1}{e}$$
and
$$f(0) = 1$$
we get that
$$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$
So for question $1$ if $g$ is an approximation of $f$ then it must be that
$$f(x)approx f(0) + xf'(0)=1+x $$
and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$
For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
$endgroup$
$begingroup$
Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
$endgroup$
– The Statistician
Jan 6 at 0:58
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
We have that
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
and so $f(x)approx f(0)+xf'(0).$ Since
$$f'(0) = frac{e-1}{e}$$
and
$$f(0) = 1$$
we get that
$$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$
So for question $1$ if $g$ is an approximation of $f$ then it must be that
$$f(x)approx f(0) + xf'(0)=1+x $$
and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$
For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
$endgroup$
$begingroup$
Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
$endgroup$
– The Statistician
Jan 6 at 0:58
add a comment |
$begingroup$
We have that
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
and so $f(x)approx f(0)+xf'(0).$ Since
$$f'(0) = frac{e-1}{e}$$
and
$$f(0) = 1$$
we get that
$$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$
So for question $1$ if $g$ is an approximation of $f$ then it must be that
$$f(x)approx f(0) + xf'(0)=1+x $$
and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$
For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
$endgroup$
$begingroup$
Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
$endgroup$
– The Statistician
Jan 6 at 0:58
add a comment |
$begingroup$
We have that
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
and so $f(x)approx f(0)+xf'(0).$ Since
$$f'(0) = frac{e-1}{e}$$
and
$$f(0) = 1$$
we get that
$$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$
So for question $1$ if $g$ is an approximation of $f$ then it must be that
$$f(x)approx f(0) + xf'(0)=1+x $$
and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$
For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
$endgroup$
We have that
$$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
and so $f(x)approx f(0)+xf'(0).$ Since
$$f'(0) = frac{e-1}{e}$$
and
$$f(0) = 1$$
we get that
$$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$
So for question $1$ if $g$ is an approximation of $f$ then it must be that
$$f(x)approx f(0) + xf'(0)=1+x $$
and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$
For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
answered Jan 6 at 0:05
Hello_WorldHello_World
4,13121731
4,13121731
$begingroup$
Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
$endgroup$
– The Statistician
Jan 6 at 0:58
add a comment |
$begingroup$
Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
$endgroup$
– The Statistician
Jan 6 at 0:58
$begingroup$
Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
$endgroup$
– The Statistician
Jan 6 at 0:58
$begingroup$
Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
$endgroup$
– The Statistician
Jan 6 at 0:58
add a comment |
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