# Thread: Limits (URGENT)

1. ## Limits (URGENT)

lim x->ln2 of (e^2x - 4) / (e^x - 2)

when ln2 is plugged in, there is an answer of 0/0 which is not what im looking for so i need to somehow factor it or something

lim x->6 of [ln(x-5)] / [ln(x-5)^3]

when 6 is plugged in, there is an answer of 0/0 which is again not what im looking for so i need to somehow factor it

2. $\displaystyle \lim_{x \to \ln 2} \frac{e^{2x} - 4}{e^{x} - 2} = \lim_{x \to \ln 2} \frac{\left(e^{x}\right)^2 - 4}{e^x - 2}$

Apply difference of squares to the numerator.

$\displaystyle \lim_{x \to 6} \frac{\ln (x - 5)}{\ln (x - 5)^3}$

Recall that $\displaystyle \ln a^b = b\ln a$. Bring down the power and something should cancel

3. Thanks!
i figured out the first one right away after you said about the difference of squares
and i brought down the 3 from the second question but im still lost from there. do i change ln to the loge? because i cant plug in 6 yet as ill still get 0/0

4. $\displaystyle \frac{\ln(x-5)}{\ln(x-5)^3} = \frac{\ln(x-5)}{3\ln(x-5)}$

What cancels?

5. yea i saw that hahah, but like, whats the point of the 6?