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

[anemone]

Here is this week's POTW:

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

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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!

 

[anemone]

Congratulations to MarkFL for his correct solution, which you can find below::)

Spoiler

Let:

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

Take the natural log of both sides:

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

Apply L'Hôpital's Rule:

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

Combine terms:

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

Apply L'Hôpital's Rule:

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

Thus:

$$L=1$$