Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

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.

share|improve this answer

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.

share|improve this answer
    
...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}$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.