# Substituting functions in limits

• Arnoldjavs3
In summary, the student was trying to solve a limit problem involving a given condition about a function. They used substitution to simplify the expression and ended up with a question about why the limit is the same when using different variables. It was explained that the variable can be named whatever is convenient and the limit will still be the same.
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
$$\lim_{x \to 1}\frac{f(x^3-1)}{x-1}.$$

## 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.

Last edited:
Arnoldjavs3 said:

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

## 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?

fresh_42 said:
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.

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
fresh_42 said:
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!

## 1. What is the definition of a limit?

A limit is the value that a function approaches as the input approaches a certain value. It is denoted by the symbol "lim" and is used to describe the behavior of a function near a specific point.

## 2. What does it mean to substitute a function in a limit?

Substituting a function in a limit means plugging in a specific value for the input variable in the given function and evaluating the resulting expression to determine the limit at that particular point.

## 3. Why is it important to understand how to substitute functions in limits?

Substituting functions in limits is important because it allows us to analyze the behavior of a function at a specific point and determine its limit. This information is crucial in understanding the overall behavior of a function and making predictions about its values.

## 4. What are some common techniques for substituting functions in limits?

Some common techniques for substituting functions in limits include direct substitution, factoring, rationalizing, and using trigonometric identities. It is important to choose the appropriate technique based on the given function and the value of the input variable.

## 5. Are there any limitations to substituting functions in limits?

Yes, there are some limitations to substituting functions in limits. It may not always give an accurate result for functions with discontinuities, such as infinite limits or limits at points where the function is undefined. In these cases, other methods, such as L'Hôpital's rule, may be used to evaluate the limit.

Replies
8
Views
648
Replies
5
Views
752
Replies
7
Views
707
Replies
17
Views
1K
Replies
8
Views
1K
Replies
13
Views
819
Replies
10
Views
1K
Replies
2
Views
503
Replies
13
Views
420
Replies
4
Views
1K