My Math Forum Please Check My Work Regarding Limit

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 February 6th, 2018, 11:02 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry 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 11:52 PM.
 February 7th, 2018, 12:45 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,974 Thanks: 1850 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*}
 February 7th, 2018, 12:47 AM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry So, the approximation of x doesn't have to be 0 to make it equals 1?
 February 7th, 2018, 12:49 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,974 Thanks: 1850 The fraction has to have the form sin(u)/u, where u tends to 0. It's okay for u to be an expression.
 February 7th, 2018, 02:30 PM #5 Global Moderator   Joined: May 2007 Posts: 6,641 Thanks: 625 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.
February 7th, 2018, 04:51 PM   #6
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Quote:
 Originally Posted by skipjack 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 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.

February 8th, 2018, 11:39 AM   #7
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Quote:
 Originally Posted by Monox D. I-Fly 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.

 February 8th, 2018, 07:28 PM #8 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Ah, I see. Thanks.

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