Question

I need help with the evaluation of this limit:

Answer 1

\[\lim_{h \rightarrow 0} \frac{ (x+h)^2\cos(x+h) -x^2\cos(x)}{ h }\]

Answer 2

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(x+h)^2\left(\sin x\cos h+\sin h\cos x\right)-x^2\cos x}{h}}\)

I would start from expanding it a bit

Answer 3

But, this is a lot to write out, what I would do is to go a regular way:

\[\lim_{h \rightarrow 0} \frac{ (x+h)^2\cos(x+h) -x^2\cos(x)}{ h }=\frac{d}{dx}\left[x^2\cos x\right]=2x\cos x-x^2\sin x \]

Answer 4

How did that all get reduced to \[\frac{d}{dx}(x^2\cos(x))\]

Answer 5

J, have you learned your derivative shortcut rules at this point?
Product rule, chain rule?

If not, then we need to do what Solomon first suggested,
expanding out all of the mess,
we'll group it in a clever way,
and relate the pieces to known limits.

Answer 6

Sorry that was directed at Jmartinez, not Johnnybean hehe

Answer 7

Hahaa

Answer 8

Oh! can't we simply substitute in 0 for x? :P

Answer 9

Recall our limit definition for a derivative,\[\large\rm \frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]So we actually have:\[\large\rm \frac{d}{dx}x^2 \cos x=\lim_{h\to0}\frac{ (x+h)^2\cos(x+h) -x^2\cos(x)}{h}\]

We clearly can't plug in h=0 right now, it's causing a problem.
Plug in x=0? Uhh.. :o

Answer 10

That gives you \(\large\rm \frac{d}{dx}x^2 \cos x\) at \(\large\rm x=0\).
But we want the general formula

Answer 11

Martinezzzzzz answer the question >.<
Have you learned your derivative shortcut rules yet?
Power rule, product rule, quotient rule, chain rule.

Or are we not up to that point in the course yet? :o
It's really important for this problem to know where you are at.

Answer 12

Ohhh... I see what he did now, @SolomonZelman reverted back to the original form and took the derivative. I got it!

Answer 13

I have learned those, yes.

Answer 14

I like chain and product better:)
All that limit mess is not for me

Answer 15

Then for this problem, unless the instructions say otherwise,
simply relate it back to the function and apply product rule! :)\[\large\rm \lim_{h\to0}\frac{ (x+h)^2\cos(x+h) -x^2\cos(x)}{h}=\frac{d}{dx}x^2 \cos x\]

Answer 16

Is that confusing? :o

Answer 17

Not really, no xD