- #1

Arnoldjavs3

- 191

- 3

## Homework Statement

I'm trying hard to understand as my professor hasn't taught(nor does my textbook) on how this works.

It is known that $$\lim_{x \to 0}\frac{f(x)}{x} = -\frac12$$

Solve

$$\lim_{x \to 1}\frac{f(x^3-1)}{x-1}.$$

## Homework Equations

## The Attempt at a Solution

OK.. so I do this:

$$\begin{align}\lim_{x \to 1}\frac{f(x^3-1)}{x-1} &= \lim_{x \to 1}\frac{f(x^3-1)}{x-1}\cdot\frac{x^2+x+1}{x^2+x+1} \\ &= \lim_{x\to 1}\frac{f(x^3-1)}{x^3-1}(x^2+x+1) \\ &= \lim_{x\to 1}\frac{f(x^3-1)}{x^3-1}\lim_{x\to 1}(x^2+x+1) \\ &= \lim_{u\to 0}\frac{f(u)}{u}(3) \end{align}$$

$$(3)(-1/2)$$

Now my question is... why is it that $$lim_{u \to 0}\frac{f(u)}{u} = \lim_{x \to 0}\frac{f(x)}{x} = -\frac12 $$

I thought that the limit changes when you substitute variables? So when we substitute ##u = x^3 -1## the limit becomes u approaches 0 as when you substitute x=1 into the above, you get u = 0. Am I wrong on this??

Note that the answer is -3/2.

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