# Thread: Why does limit tend to positve infinity and 0?

1. ## Why does limit tend to positve infinity and 0?

Hi everyone,

I have following task: When I approach from the right, it's positive infinity (which is correct) but why is it 0 when I approach from the left, why isn't it negative infinity?

Thank you in advance!

Kind regards!

2. ## Re: Why does limit tend to positve infinity and 0? Originally Posted by ORT Hi everyone,
I have following task: What task? When I approach from the right, it's positive infinity (which is correct) but why is it 0 when I approach from the left, why isn't it negative infinity?
Thank you in advance!
You have posted a simple statement. What is the question?

3. ## Re: Why does limit tend to positve infinity and 0?

Your function is $\displaystyle e^{\frac{1}{x-1}}$ and you are asking about the limit as x goes to 1 from above and below. As x approaches 1 from above, x- 1 is positive so we have e to a larger and larger positive value. As x approaches 1 from below, x- 1 is negative so we have e to a "larger and larger" negative power. Letting y= 1/(x- 1), the limit as x approaches 1 from below is the same as $\displaystyle \lim_{y\to -\infty} e^y$. And that limit is 0.

4. ## Re: Why does limit tend to positve infinity and 0? Originally Posted by HallsofIvy Your function is $\displaystyle e^{\frac{1}{x-1}}$ and you are asking about the limit as x goes to 1 from above and below. As x approaches 1 from above, x- 1 is positive so we have e to a larger and larger positive value. As x approaches 1 from below, x- 1 is negative so we have e to a "larger and larger" negative power. Letting y= 1/(x- 1), the limit as x approaches 1 from below is the same as $\displaystyle \lim_{y\to -\infty} e^y$. And that limit is 0.
I wish the we did not have make a guess as to exact statement of a question.

5. ## Re: Why does limit tend to positve infinity and 0?

If you want to see clearly why coming from the Left to 1 tends to zero, do this

take a number less than 1, but almost equal to 1
for example 0.9

now substitute in 1/(x - 1)
when x = 0.9, 1/(x - 1) = 1/(0.9 - 1) = 1 / -0.1 = -10

now e^(-10) = 0.000045 (which almost zero)

now take x = 0.99

after substitution
e^(-100) = 0.000000000000000000000000000000000000000000037 (wOW it is so clear as x goes to 1 from the left the function becomes zero)

try x = 0.999 if you want

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