# Thread: 2 Questions on L-Hopitals Rule

1. ## 2 Questions on L-Hopitals Rule

I have one question that says

lim x-> (3^x - 1) / (2^x - 1)

lim x -> 0 (ln3 x 3^x) / (ln2 x 2^x)

Is that correct? Where would I go from there?

Also I have a question

lim y->0 ((sqroot 5y+25) - 5) / y

I am not sure how I would do this problem.

Thanks for the help!

2. Originally Posted by KarlosK
I have one question that says

lim x-> (3^x - 1) / (2^x - 1)

lim x -> 0 (ln3 x 3^x) / (ln2 x 2^x)

Is that correct? Where would I go from there?
Substitute x = 0?

Also I have a question

lim y->0 ((sqroot 5y+25) - 5) / y

I am not sure how I would do this problem.

Thanks for the help!
Are you forced to use L'Hospital's rule??

Without L'Hospital's you can just rationalize the numerator for a simple solution.

3. Note that the second limit is the derivative of $\sqrt{5x}$ at $x=5$.

4. It says solve using l hopitals rule. That is what we are covering in class. Our teacher does not teach us very well though

5. Originally Posted by Isomorphism
Substitute x = 0?

Are you forced to use L'Hospital's rule??

Without L'Hospital's you can just rationalize the numerator for a simple solution.
If i substitute in X wont I just wind up with 0 over 0 again?

6. He meant to substitute x=0 in the formula you got: $\lim_{x\to 0} {\ln3 \cdot 3^x\over \ln2 \cdot 2^x}$.

He meant to substitute x=0 in the formula you got: $\lim_{x\to 0} {\ln3 \cdot 3^x\over \ln2 \cdot 2^x}$.
8. $3^0=0$? Are you quite sure? Doesn't that mean that $0 = 0\cdot 3^1 = 3^0\cdot3^1 = 3^{0+1} = 3^1 = 3$?