# Substituting functions in limits

1. Dec 12, 2016

### Arnoldjavs3

1. The problem statement, all variables and given/known data
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
$$\lim_{x \to 1}\frac{f(x^3-1)}{x-1}.$$

2. Relevant equations

3. 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 in to the above, you get u = 0. Am I wrong on this??

Note that the answer is -3/2.

Last edited: Dec 12, 2016
2. Dec 12, 2016

### Staff: Mentor

You may name the variable whatever you like:
Whether it's $\lim_{u \to 0}\frac{f(u)}{u}\;$ or $\; \lim_{\text{ tree} \to 0}\frac{f(\text{tree})}{\text{tree}}\;$ or $\;\lim_{x \to 0}\frac{f(x)}{x}$

Or wasn't that the question and it's about the previous steps?

3. Dec 12, 2016

### Arnoldjavs3

I was asking why is it that lim u-> 0 f(u)/u = lim x -> 0 f(x)/x in the last step of the solution.

4. Dec 12, 2016

### Staff: Mentor

It is simply the same, identical. First, $u$ is simply an abbreviation of $x^3-1$. But then we forget about the origin of $u$, because it doesn't matter anymore. Once we have it, we can concentrate on the expression $\lim_{u \to 0}\frac{f(u)}{u}$ and we are free to rename it.
If it feels better for you, then how about changing the given condition for $f$ to $\lim_{v \to 0}\frac{f(v)}{v}=-\frac{1}{2}$ and substitute another time $v=u$ at the end of the proof?

5. Dec 12, 2016

### Arnoldjavs3

If it is the same, then that's enough for me to know. Thanks!