Substituting functions in limits

[Arnoldjavs3]

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.

 

[fresh_42]

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

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.

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?

 

[Arnoldjavs3]

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?

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.

 

[fresh_42]

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?

 

[Arnoldjavs3]

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?

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