Question

Evaluate the following limit:
\[\lim_{x \rightarrow 0-} \frac{e^{1/x}}{x}\]

Answer 1

@phi

Answer 2

@amistre64
@AccessDenied
@JamesJ
@SmoothMath (this is a different question :p)

Answer 3

I know this is easy, but I do not know how to proceed with this

Answer 4

I believe L'hopital's rule applies... but I'm not sure if it's usable for one sided limits.

Answer 5

It is

Answer 6

Well then cool. Do you see that the top and bottom both approach 0?

Answer 7

From the left, that is. From the right, the top would approach infinity.

Answer 8

How will it work? You get:
\[\lim_{x \rightarrow 0-} \frac{e^{1/x}*-1/x^2}{1}\]

Answer 9

Which will then go to the denominator, and then you do it and over and over and over

Answer 10

I think there is no need for L'H Rule here

Answer 11

I think you're right.

Answer 12

But then wait, if I sub in close to 0, then x becomes 0, but then e^{1/x} goes to infinity (v.large number) that goes into the denominator

Answer 13

so we get:
\[\frac{1}{\infty*0}\]

Answer 14

You have a very large number multiplied by 0. That is 0 then?

Answer 15

\[\infty * 0\]

Answer 16

b/c:
\[\frac{1}{\infty}*\frac{1}{0}\]

Answer 17

No no. It's still as x approaches 0 from the left, which means e^(1/x) approaches 0.

Answer 18

i dont follow the argument

Answer 19

must be zero ....
exponential function always win over polynomial function ...

Answer 20

let \(u=-\frac{1}{x}\)

then \[\lim_{x \to 0^{-}} \frac{e^{1/x}}{x}=\lim_{u\to\infty}-\frac{e^{-u}}{1/u}\]

\[=-\lim_{u\to\infty}\frac{u}{e^{u}}\]

now apply L'Hospital's rule

Answer 21

Seems legit.

Answer 22

the answer is \(0^-\)right?

Answer 23

or just \(0\)

Answer 24

the answer is zero

Answer 25

And to test that, you can check values increasingly closer to 0 from the left. Now, this isn't fool-proof. Some functions will mislead you if you try this kind of test. It usually works though.