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February 6th, 2018, 10:02 PM   #1
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Please Check My Work Regarding Limit

The value of $\displaystyle \lim_{x\to\frac{1}{2}\pi}\frac{1-\sin x}{x-\frac{1}{2}\pi}$ is ....
a. -2
b. -1
c. 1
d. 0
e. 2

What I did:
$\displaystyle \lim_{x\to\frac{1}{2}\pi}\frac{1-\sin x}{x-\frac{1}{2}\pi}$
$\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{\sin\frac{1}{2}\pi-\sin x}{x-\frac{1}{2}\pi}$
$\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}\sin\frac{\frac{1}{2}\pi-x}{2}}{-(\frac{1}{2}\pi-x)}$
$\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}\sin\frac{\frac{1}{2}\pi-x}{2}}{-(\frac{1}{2}\pi-x)}\cdot\frac{\frac{1}{2}}{\frac{1}{2}}$
$\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}\sin\frac{\frac{1}{2}\pi-x}{2}}{-2\frac{(\frac{1}{2}\pi-x)}{2}}$
$\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}}{-2}\cdot\frac{\sin\frac{\frac{1}{2}\pi-x}{2}}{\frac{(\frac{1}{2}\pi-x)}{2}}$
$\displaystyle =\lim_{x\to\frac{1}{2}\pi}(-\cos\frac{\frac{1}{2}\pi+x}{2})\cdot\frac{\sin \frac{\frac{1}{2}\pi-x}{2}}{\frac{(\frac{1}{2}\pi-x)}{2}}$
From then on is what got me hesitated. Does $\displaystyle \lim_{x\to\frac{1}{2}\pi}\frac{\sin\frac{\frac{1}{ 2}\pi-x}{2}}{\frac{(\frac{1}{2}\pi-x)}{2}}$ equal 1? Because as far as I recall, this kind of things can only work if x approaches 0, can't it? However, if I keep treating $\displaystyle \lim_{x\to\frac{1}{2}\pi}\frac{\sin\frac{\frac{1}{ 2}\pi-x}{2}}{\frac{(\frac{1}{2}\pi-x)}{2}}$ as 1, I got this:
$\displaystyle \lim_{x\to\frac{1}{2}\pi}(-\cos\frac{\frac{1}{2}\pi+x}{2})\cdot\frac{\sin \frac{\frac{1}{2}\pi-x}{2}}{\frac{(\frac{1}{2}\pi-x)}{2}}$
$\displaystyle =-\cos\frac{\frac{1}{2}\pi+\frac{1}{2}\pi}{2})\cdot1$
$\displaystyle =-\cos\frac{\pi}{2}=-0=0$
Please tell me how to solve it properly.

Last edited by skipjack; February 6th, 2018 at 10:52 PM.
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February 6th, 2018, 11:45 PM   #2
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It's correct.

$\displaystyle \begin{align*}\lim_{x\to\frac12\pi}\frac{1 - \sin x}{x - \frac12\pi} &= \lim_{x\to\frac12\pi}\frac{1 - \cos\left(x - \frac12\pi\right)}{x - \frac12\pi} \\
&= \lim_{x\to\frac12\pi}\frac{2\sin^2{\Large(}\frac{x \, - \frac12\pi}{2}{\Large)}}{x - \frac12\pi}\cdot\frac{\frac12}{\frac12} \\
&= \lim_{x\to\frac12\pi}\sin{\huge(}\frac{x - \frac12\pi}{2}{\huge)}\cdot\frac{\sin{\Large(} \frac{x\, - \frac12\pi}{2}{\Large)}}{\frac{x\, - \frac12\pi}{2}} \\
&= 0(1) \\
&= 0 \text{ (choice d.)}\end{align*}$
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February 6th, 2018, 11:47 PM   #3
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So, the approximation of x doesn't have to be 0 to make it equals 1?
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February 6th, 2018, 11:49 PM   #4
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The fraction has to have the form sin(u)/u, where u tends to 0.

It's okay for u to be an expression.
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February 7th, 2018, 01:30 PM   #5
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Are you allowed to use L'hopital's rule? In that case the limit becomes a question of limit for -cos(x)/1, which =0.
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February 7th, 2018, 03:51 PM   #6
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Quote:
Originally Posted by skipjack View Post
The fraction has to have the form sin(u)/u, where u tends to 0.

It's okay for u to be an expression.
So it has nothing to do with whether x approaches 0 or not?

Quote:
Originally Posted by mathman View Post
Are you allowed to use L'hopital's rule? In that case the limit becomes a question of limit for -cos(x)/1, which =0.
No, because the chapter discussing about derivation will come right after this limit chapter.
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February 8th, 2018, 10:39 AM   #7
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Quote:
Originally Posted by Monox D. I-Fly View Post
So it has nothing to do with whether x approaches 0 or not?
In this example, what counts is that $x$ approaches $\pi/2$, as that implies that $\displaystyle \frac{x\, - \frac12\pi}{2}$ approaches zero.
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February 8th, 2018, 06:28 PM   #8
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Ah, I see. Thanks.
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