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February 6th, 2018, 10:02 PM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 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: $\displaystyle \lim_{x\to\frac{1}{2}\pi}\frac{1\sin x}{x\frac{1}{2}\pi}$ $\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{\sin\frac{1}{2}\pi\sin x}{x\frac{1}{2}\pi}$ $\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}\sin\frac{\frac{1}{2}\pix}{2}}{(\frac{1}{2}\pix)}$ $\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}\sin\frac{\frac{1}{2}\pix}{2}}{(\frac{1}{2}\pix)}\cdot\frac{\frac{1}{2}}{\frac{1}{2}}$ $\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}\sin\frac{\frac{1}{2}\pix}{2}}{2\frac{(\frac{1}{2}\pix)}{2}}$ $\displaystyle =\lim_{x\to\frac{1}{2}\pi}\frac{2\cos\frac{\frac{1 }{2}\pi+x}{2}}{2}\cdot\frac{\sin\frac{\frac{1}{2}\pix}{2}}{\frac{(\frac{1}{2}\pix)}{2}}$ $\displaystyle =\lim_{x\to\frac{1}{2}\pi}(\cos\frac{\frac{1}{2}\pi+x}{2})\cdot\frac{\sin \frac{\frac{1}{2}\pix}{2}}{\frac{(\frac{1}{2}\pix)}{2}}$ From then on is what got me hesitated. Does $\displaystyle \lim_{x\to\frac{1}{2}\pi}\frac{\sin\frac{\frac{1}{ 2}\pix}{2}}{\frac{(\frac{1}{2}\pix)}{2}}$ 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 $\displaystyle \lim_{x\to\frac{1}{2}\pi}\frac{\sin\frac{\frac{1}{ 2}\pix}{2}}{\frac{(\frac{1}{2}\pix)}{2}}$ as 1, I got this: $\displaystyle \lim_{x\to\frac{1}{2}\pi}(\cos\frac{\frac{1}{2}\pi+x}{2})\cdot\frac{\sin \frac{\frac{1}{2}\pix}{2}}{\frac{(\frac{1}{2}\pix)}{2}}$ $\displaystyle =\cos\frac{\frac{1}{2}\pi+\frac{1}{2}\pi}{2})\cdot1$ $\displaystyle =\cos\frac{\pi}{2}=0=0$ Please tell me how to solve it properly. Last edited by skipjack; February 6th, 2018 at 10:52 PM. 
February 6th, 2018, 11:45 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,535 Thanks: 1750 
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 6th, 2018, 11:47 PM  #3 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
So, the approximation of x doesn't have to be 0 to make it equals 1?

February 6th, 2018, 11:49 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,535 Thanks: 1750 
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, 01:30 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,586 Thanks: 612 
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, 03:51 PM  #6  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry  Quote:
No, because the chapter discussing about derivation will come right after this limit chapter.  
February 8th, 2018, 10:39 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,535 Thanks: 1750  
February 8th, 2018, 06:28 PM  #8 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
Ah, I see. Thanks.


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