How Do You Approach the Limit of π/2 Using Trigonometric Functions?

[twoflower]

Hi all, I've been practising limits for test, which is approaching very fast :) Let's have this limit:

<br /> \lim_{x \rightarrow \frac{\pi}{2}_{+}} \frac{ \sqrt{1 - \sin x}}{\log \frac{2x}{ \pi }}<br />

I played with that so I got this form:

<br /> \lim \frac{1}{\sqrt{1+\sin x}} \frac{1}{\frac{\log \frac{2x}{\pi}}{\cos x}}<br />

It can be seen that the second fraction is inversed well known limit

<br /> \lim_{x \rightarrow 1} \frac{ \log{x} } { x - 1 } = 1<br />

Then, I can write

<br /> = \lim \frac{1}{ \sqrt{1 + \sin x }} = \frac{ \sqrt{2} } { 2 }<br />

The problem is, I don't know how to justify my method exactly. You see, we have to base all your steps on the theorem we use for it, we must prove the conditions and so on...And in this limit I did it rather ty intuition...

So how would the precise solution look like?

Thank you.

 

[HallsofIvy]

I don't see any reason to make it so complex. AT x= &pi; sin x= 0 so the numerator is 1. Assuming that that "3.14" in the denominator is actually &pi;, then the denominator is log 2 which is not 0. Since all functions are continuous in the appropriate neighbohoods, the limit is 1/log(2).

 

[twoflower]

I don't see any reason to make it so complex. AT x= π sin x= 0 so the numerator is 1. Assuming that that "3.14" in the denominator is actually π, then the denominator is log 2 which is not 0. Since all functions are continuous in the appropriate neighbohoods, the limit is 1/log(2).

I'm sorry, I didn't write the LaTeX code properly, so I had to edit it. Now it's how it should be.