# Why lim x->0 e^-x - 1 / x = -1

## Homework Statement

In the textbook I am studying one of the exercises is: lim x→0 e-x - 1 / x = ? The answer given is -1. But I can't see why this answer should follow.

## Homework Equations

The most relevant equation here is: lim x→0 = ex - 1 / x = 1

## The Attempt at a Solution

So let's take: lim x→0 e-x - 1 / x = ? So x approximates 0, so you get e-x approximates e-0 which in my view is just e0 = 1. Therefore I don't see how this differs from lim x→0 = ex - 1 / x = 1. In my view the answer should be 1 and not -1... I'm clearly missing something, but I don't have a clue what... Can you please help me?

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Mark44
Mentor

## Homework Statement

In the textbook I am studying one of the exercises is: lim x→0 e-x - 1 / x = ? The answer given is -1. But I can't see why this answer should follow.
You really need parenthese.
Stoney Pete said:
e-x - 1 / x means ##e^{-x} - \frac 1 x##, which is probably not what you intended. If not using Tex, write this as (e-x - 1)/x.

## Homework Equations

The most relevant equation here is: lim x→0 = ex - 1 / x = 1

## The Attempt at a Solution

So let's take: lim x→0 e-x - 1 / x = ? So x approximates 0, so you get e-x approximates e-0 which in my view is just e0 = 1. Therefore I don't see how this differs from lim x→0 = ex - 1 / x = 1.
Again, you need parentheses.
The problem with your analysis is that both the numerator and denominator are approaching 0,
Stoney Pete said:
In my view the answer should be 1 and not -1... I'm clearly missing something, but I don't have a clue what... Can you please help me?
Your textbook must have examples of problems where both numerator and denominator are approaching 0. Have you looked at them?

BTW, I agree with the textbook's answer.

Mark44
Mentor
Thread moved from the Precalc section. Problems about limits are typically at the Calculus level, not the Precalc level.

Thanks for the tip about the parentheses....

The problem with the text book I am studying is that it can be a bit lacking in explanation and illustration sometimes.... It's a very comprehensive text book, but also overly concise sometimes...

Anyway, I appreciate the problem that the denominator is approaching zero... You can't divide through zero... So I probably have to rework the division in such a way that the denominator no longer approximates zero.... I probably have to get back to the chapter on limits, wich I found somewhat hard to follow near the end... I probably missed something there which I need to use here.... But if you could just give me hint in which direction I must look for the answer it would be much appreciated!

PeroK
Homework Helper
Gold Member
Thanks for the tip about the parentheses....

The problem with the text book I am studying is that it can be a bit lacking in explanation and illustration sometimes.... It's a very comprehensive text book, but also overly concise sometimes...

Anyway, I appreciate the problem that the denominator is approaching zero... You can't divide through zero... So I probably have to rework the division in such a way that the denominator no longer approximates zero.... I probably have to get back to the chapter on limits, wich I found somewhat hard to follow near the end... I probably missed something there which I need to use here.... But if you could just give me hint in which direction I must look for the answer it would be much appreciated!
Three suggestions:

1) Plug in a small value of x into a calculator or spreadsheet and see whether the function is close to ##1## or ##-1##.

2) What definitions of ##e^{-x}## do you know?

3) Do you think this might be a "hospital" case?

cnh1995
Thanks for the tips... I will get to work... Sometimes I really hate that I love mathematics while sucking at it :)

Mark44
Mentor
I probably have to get back to the chapter on limits, wich I found somewhat hard to follow near the end... I probably missed something there which I need to use here
Yes, good idea.

One thing that you seem to be missing is that a limit expression of the form ##[\frac 0 0]## is one of several indeterminate forms. It is indeterminate because the limit itself can be any number, depending on the expression itself.

Here are three examples where both the numerator and denominator are approaching zero, but with wildly different limiting values.
1. ##\lim_{x \to 1} \frac {x^2 + 3x - 4}{x - 1}## (The limit is 5.)
2. ##\lim_{x \to 3} \frac {x^2 - 6x + 9}{x - 3}## (The limit is 0.)
3. ##\lim_{x \to -2} \frac {x + 2}{x^3 - 6x^2 + 12x + 8}## (The limit is ∞.)

cnh1995

2. Homework Equations

The most relevant equation here is: lim x→0 = ex - 1 / x = 1
$$\lim_{x \to 0} \frac {e^x-1}{x }$$ is a special case of $$\lim_{x \to 0} \frac {a^x-1}{x }$$