Approaching the Limit of a Sin Function Using Calculus Techniques

Jimbo57
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Homework Statement



[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP120619i0ea2h15bci31300005d9a3422292h43h4?MSPStoreType=image/gif&s=43&w=174&h=39

Homework Equations


The Attempt at a Solution



So Wolfram says to use L'Hopital as the first step, we haven't learned anything about this yet so there has to be another way using calculus that a first year math student would know.

I would first expand the denominator of x^2-4 to (x+2)(x-2) and cancel out the x+2 in both numerator and denominator. And I'm left with:
((x-1)sin1)/(x-2)
This is where I get lost...

Any help would be much appreciated!
Jim
 
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One cannot separate (x+2) from sin any more than one can separate ... a bone from the mouth of a very hungry dog!

Either one 'cancels' the whole sin(x+2) or nothing at all.
 
You don't cancel the (x+2)'s. You should know

\lim_{x\rightarrow 0}\frac{\sin x}{x}

Use that somehow.
 
Did you mention in class that lim(a x b) = lim(a) x lim(b)?
 
Yes, we mentioned all the above, and this was much easier after LCKurtz's suggestion:

=lim sin((x+2)/(x+2)) * lim (x-1)/(x-2)
= 1 * -3/-4
=3/4
I really need to learn how to use that equation generator thing...

Much appreciated folks! :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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