Question

What is the limit of x as x approaches 0 from both sides. When (1/e^x) - (1/x)?

Answer 1

you want to know\[\lim_{x\to0^-}\left(\frac1{e^x}-\frac1x\right)\]and\[\lim_{x\to0^+}\left(\frac1{e^x}-\frac1x\right)\]?

Answer 2

Its approaching from both sides, that's the question, I'm assuming it doesn't exist though, I just wanted some feedback.

Answer 3

no, this limit will diverge from both sides

Answer 4

I forgot to state we're using the L'Hopital rule, sorry

Answer 5

\[\lim_{x\to0}\left(\frac1{e^x}-\frac1x\right)\]then we need to turn it into an indeterminate \(\frac00\) form first..\[\lim_{x\to0}\left(x-e^x\over xe^x\right)\]

Answer 6

oh that's not indeterminate \(\frac00\) though :/

Answer 7

Could you explain the indeterminate form and how i go about converting it? That's really my problem here.

Answer 8

we can't use l'hospital until we get it so that whne you plug in the number that x is approaching we get either\[\frac\infty\infty\text{ or }\frac00\]so sometimes we have to manipulate an expression to get it into that form

Answer 9

for example\[\lim_{x\to0^+}x^2\ln x\]if we plug in 0 right away we get\[0\cdot\ln0=0\cdot(-\infty)\]which is undefined
but if we do a little rearranging...\[\lim_{x\to0^+}x^2\ln x=\lim_{x\to0^+}\frac{\ln x}{\frac1{x^2}}\]plugging in gives\[\frac{\ln0}{\frac10}=-\frac\infty\infty\]so after that little algebra trick we can use l'hospital to find the limit, because nopw it is in one of our two possible indeterminate forms (note that \(\pm\) does not matter)

Answer 10

I understand it now, thanks.

Answer 11

welcome, glad to help :)