Question

Can someone help me with L'Hospital's Rule consisting with limits?

what is the limit of [x/(x-1)]-[1/lnx] as x approaches 1? Thank you!

Answer 1

Ok I found a solution to this. It's a bit long to type so bear with me

Answer 2

okay! :) thank you!

Answer 3

HR in the following is for L'Hôpital's Rule
\[\large \lim_{x\rightarrow 1 }\left( \frac{x}{x-1}-\frac{1}{\ln x}\right)\\
=\large \lim_{x\rightarrow 1}\left( \frac{x(\ln x)}{(x-1)(\ln x)}-\frac{(x-1)}{\ln x(x-1)}\right), \text{put same denominator}\\
=\large \lim_{x\rightarrow 1}\left( \frac{x\ln x -x +1}{(x-1)\ln x}\right),\text{form }\frac{0}{0}\implies \text{use l'HR}\\
=\large \lim_{x\rightarrow 1}\left( \frac{\ln x+x\frac{1}{x}-1}{\ln x+(x-1)\frac{1}{x}}\right)\\
=\large \lim_{x\rightarrow 1}\left( \frac{\ln x}{\ln x+1-\frac{1}{x}}\right)\text{form }\frac{0}{0}\implies \text{use l'HR}\\
=\large \lim_{x\rightarrow 1}\left( \frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^2}}\right)\\
=\frac{\frac{1}{1}}{\frac{1}{1}+\frac{1}{1^2}}\\
=\frac{1}{2}\]

Answer 4

Okay, I see where I made my mistake now. I didn't take the derivative properly. Thank you!!!! You're awesome.

Answer 5

:)