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

[Stoney Pete]

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 = e^x - 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 = e^x - 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?

 

[Mark44]

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.

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 = e^x - 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 = e^x - 1 / x = 1.

Again, you need parentheses.
The problem with your analysis is that both the numerator and denominator are approaching 0,

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]

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

 

[Stoney Pete]

Thanks for the tip about the parentheses...

The problem with the textbook I am studying is that it can be a bit lacking in explanation and illustration sometimes... It's a very comprehensive textbook, 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, which 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]

Thanks for the tip about the parentheses...

The problem with the textbook I am studying is that it can be a bit lacking in explanation and illustration sometimes... It's a very comprehensive textbook, 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, which 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?

 

[Stoney Pete]

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

 

[Mark44]

I probably have to get back to the chapter on limits, which 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 ∞.)

 

[Vishakha]

2. Homework Equations
The most relevant equation here is: lim x→0 = e^x - 1 / x = 1

$$\lim_{x \to 0} \frac {e^x-1}{x }$$ is a special case of $$\lim_{x \to 0} \frac {a^x-1}{x }$$