Question

limit as x approaches 0 of ((2-h)^5-32)/(h)

Answer 1

Just wondering - do you know how to compute derivatives? Or are you just learning how to do that and have to rely on the limit definition?

Answer 2

I know derivatives

Answer 3

I like derivatives better. :) haha

Answer 4

Ok, that makes things easier then.

Note that \(\large \displaystyle\lim_{h\to 0}\frac{(2-h)^5-32}{h}\) looks something like \(\large\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=f^{\prime}(a)\); in particular, \[\large\lim_{h\to 0}\frac{(2-h)^5-32}{h} = -\lim_{h\to 0}\frac{((-2)+h)^5 -(-32)}{h}=\ldots\]Can you take things from here? :-)

Answer 5

wait so I can just find the derivative?

Answer 6

Yup, that's the trick here. XD

But the derivative of what function at what point?

Answer 7

Hmm I am not sure

Answer 8

So I take the derivative and then evaluate that limit at 0

Answer 9

Would it be clearer if I told you that \[\large \lim_{h\to 0}\frac{(2-h)^5-32}{h}= \lim_{h\to 0}\frac{[-((-2)+h)^5] - [-(-2)^5]}{h}?\]

Answer 10

so then i evaluate it as h approaches zero. Or do I not do that?

Answer 11

The point of this "trick" is to avoid computing a limit completely and instead evaluate a derivative at a point.

Answer 12

The observation to be made here is that the limit your given looks like the difference quotient of some function at a certain point. The objective then is to determine what function you're differentiating and at what point.

Answer 13

There's nothing wrong with going ahead and expanding everything out and then simplify the whole thing; it's just that this method saves you a lot of time, especially if you're familiar with derivatives at this point.