Question

L'Hospital's Rule...

Answer 1

Sure, so how far can you get on your own with either of these?

Answer 2

I know how to apply Bernoulli's rule, but only if the function is given as a fraction. Do I rationalize to make it so?

Answer 3

I have the answers already, actually. I'm just double-checking to see if they're right. The answer for letter a is 2 and the answer for letter b is 1. I would like to compare solutions with someone else's work.

Answer 4

Sure, so what I would do is just multiply by the conjugate for the first one, \[\frac{\sqrt{x^2+x}+\sqrt{x^2-x}}{\sqrt{x^2+x}+\sqrt{x^2-x}}\] if that's what you did, you probably got the right answer. For the second one I'd use exponent rules to turn f(x) into e^ln(f(x)) if that makes sense.

Answer 5

To just check answers directly, I suggest just typing them into wolfram alpha.

Answer 6

Wolfram Alpha doesn't seem to make sense of L'Hospital's Rule...

Okay. First one is good.

Second one, yeah, that's what I did, but I was hoping to go further than that.

Thanks.

Answer 7

\[\lim_{x \rightarrow 1^+}(x-1)^{\ln x}=\lim_{x \rightarrow 1^+}e^{\ln[ (x-1)^{\ln x}]}\]
Soall I did was take the natural log and then exponetiate it. It doesn't change it because it is like taking the square root and then squaring it again.\[\lim_{x \rightarrow 1^+}e^{\ln[ (x-1)^{\ln x}]}=e^{\lim_{x \rightarrow 1^+}{\ln x *\ln (x-1)}}\] Here I used exponent rules to take the lnx down to multiply out front. and I also moved the limit up into the exponent of e since e is just a constant, the limit doesn't depend on it right? I'll leave it out and we'll exponentiate our answer when we're done.
\[\lim_{x \rightarrow 1^+} \frac{\ln (x-1)}{1/\ln x}\] All I did was rewrite it so that now we have it in a form that works. When we plug in 1 we'll get ln0/(1/ln1) which is the same as infinity/infinity. Now you can L'H-ize it up.