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I'm asked to find the following limit:

$\lim\limits_{x\rightarrow 0} \frac{\sin(3+x)^2-\sin(9)}{x}$

I easily found the solution using L'Hopital's rule as follow:

$\lim\limits_{x\rightarrow 0} \frac{\sin(3+x)^2-\sin(9)}{x}=\lim\limits_{x\rightarrow 0} (6+2x)\cos(3+x)^2=6 \cos(9)=-5,46$

I double-checked my result graphing my function and it works. The problem is that I'm not supposed to know L'Hopital's rule at this time. Is there an alternative way of finding this limit?

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up vote 5 down vote accepted

The limit is by definition the derivative of $\sin(t^2)$ at $t=3$. Just replace $x$ by $h$, and you will see it.

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You can use the difference-quotient definition of the derivative, $$f'(a)=\lim_{h\to0}{f(a+h)-f(a)\over h}$$ if you just choose $f$, $a$, and $h$ appropriately.

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...assuming it is $\sin((x+3)^2)$ and not $(\sin(x+3))^2$. – GEdgar Apr 4 '12 at 13:23

$\frac{(\sin (x+3)^2 - \sin 9)}{(x+3)^2 -9} \frac{(x+3)^2 - 9}{x}$ Let x tend to zero. Then $6 \cos{9}$ is the answer. Note that the first quantity is as follows: $ \lim_{y -> 9} \frac{\sin y - \sin{9}}{y-9} = \cos{9}$

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This is the differentiation of the composite function, $f,g$ where $f(x)=\sin x, g(x)=(x+3)^2$ – hkju Apr 4 '12 at 13:53
$\frac{d}{dx}(f(g(x))$ – hkju Apr 4 '12 at 20:40

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