Evaluating a Limit: Examining lim h->0 ((8+h)^⅓ -2)/h

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Homework Statement


Evaluate lim h->0 ((8+h)^⅓ -2)/h.

Homework Equations


Hint: Let 8+h=x^3

The Attempt at a Solution


I've uploaded a picture of my calculation. But I am not sure if that is the final answer or is there a following step to get the answer.
 

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Why do you think that as h->0 with x^3=8+h that x->(-3)? And why do you think that (x^3)^(1/3)=x^3?
 
Aviegaille said:

Homework Statement


Evaluate lim h->0 ((8+h)^⅓ -2)/h.

Homework Equations


Hint: Let 8+h=x^3

The Attempt at a Solution


I've uploaded a picture of my calculation. But I am not sure if that is the final answer or is there a following step to get the answer.

Please do not post thumbnails; they cannot be viewed on some media! Just type out things directly.
 
I don't know exactly which of many possible methods you are expected to use here, but did you notice that this limit is precisely that defining the derivative of f(x) at x= 8, with f(x)= x^{2/3}?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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