What is the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0?

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  • Thread starter anemone
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    2016
In summary, the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0 is equal to 1. To solve for this limit, we can use the power rule and the limit of trigonometric functions. This is because as x approaches 0, the expression (1 + tanx) approaches 1 and (1 + sinx)^(1/x^2) approaches 1, resulting in a limit of 1/1 which is equal to 1. L'Hopital's rule cannot be used to solve for this limit as the expression is not in an indeterminate form. The limit of this expression as x approaches 0 also relates to the graph of
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anemone
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Here is this week's POTW:

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Evaluate \(\displaystyle \lim_{{x}\to{0}}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{x^2}}\).

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to MarkFL for his correct solution, which you can find below::)

Let:

\(\displaystyle L=\lim_{x\to0}\left(\frac{1+\tan(x)}{1+\sin(x)}\right)^{\frac{1}{x^2}}\)

Take the natural log of both sides:

\(\displaystyle \ln(L)=\lim_{x\to0}\left(\frac{\ln(1+\tan(x))}{x^2}-\frac{\ln(1+\sin(x))}{x^2}\right)\)

Apply L'Hôpital's Rule:

\(\displaystyle \ln(L)=\frac{1}{2}\lim_{x\to0}\left(\frac{\sec^2(x)}{x(1+\tan(x)))}-\frac{\cos(x)}{x(1+\sin(x))}\right)\)

Combine terms:

\(\displaystyle \ln(L)=\frac{1}{2}\lim_{x\to0}\left(\frac{\sec^2(x)(1+\sin(x))-\cos(x)(1+\tan(x))}{x(1+\tan(x))(1+\sin(x))}\right)\)

Apply L'Hôpital's Rule:

\(\displaystyle \ln(L)=\frac{1}{2}\lim_{x\to0}\left(\frac{\sin(x)(\tan(x)+1)+2(\sin(x)+1)\tan(x)\sec^2(x)}{(\sin(x)+1)(\tan(x)+1)+x\cos(x)(\tan(x)+1)+x(\sin(x)+1)\sec^2(x)}\right)=0\)

Thus:

\(\displaystyle L=1\)
 

FAQ: What is the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0?

What is the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0?

The limit of this expression as x approaches 0 is equal to 1.

How do you solve for the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0?

To solve for this limit, we can use the power rule and the limit of trigonometric functions to simplify the expression and then evaluate the limit.

Why is the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0 equal to 1?

This is because as x approaches 0, the expression (1 + tanx) approaches 1 and (1 + sinx)^(1/x^2) approaches 1, resulting in a limit of 1/1 which is equal to 1.

Can you use L'Hopital's rule to solve for the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0?

No, L'Hopital's rule cannot be used as the expression is not in an indeterminate form.

How does the limit of (1 + tanx)/(1 + sinx)^(1/x^2) as x approaches 0 relate to the graph of the function?

The limit of this expression as x approaches 0 is equal to the vertical asymptote of the function, which is located at x=0 on the graph.

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