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 February 8th, 2013, 12:04 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Can I apply L'hopital here? hi all $\lim {x \to \0^+} x^x= A$ $lnA=\lim {x \to \0^+} xlnx = \lim {x \to \0^+} \frac{lnx}{\frac{1}{x}} = {\frac{-\infty}{\infty}}$ Can i apply lhopital when the limit of the numerator is different than the denominator but both are infinite?
 February 8th, 2013, 01:24 AM #2 Member     Joined: Jul 2012 Posts: 60 Thanks: 0 Math Focus: Calculus Re: Can I apply L'hopital here? I think so. However, I've been wrong about this before so I'm going to give you an alternative here. My teacher never taught us Lhopitals rule. She took a different approach. She drew the graphs of the following functions: y=ln(x); y=sqrt(x); y=x^2; y=e^x as an example. Then we looked at the graphs and saw that as x aproaches infinity all these functions approach infinity. But ln(x) approaches infinity the slowest. If you draw it you will see the function (y=ln(x)) grows very slowly compared to y=sqrt(x). So y=ln(x) approaches infinity a lot slower than y = sqrt(x) So we ranked these functions based on how fast they grow when x->infinity. y=ln(x) is the slowest, followed by y=sqrt(x) then y=x, after that goes y=x^2 (or x^n) and the fastest growing function is y=e^x (or a^x where a>1) But this only applies when x->infinity, not when x->0 or any other number. $\ln A=\lim_{x\to 0^+}\frac{\ln x}{\frac{1}{x}}$ we can say 1/x = t, so the the limit becomes the following: $\ln A=\lim_{t\to\infty}\frac{\ln \frac{1}{t}}{t}$ $\ln A=\lim_{t\to\infty}\frac{- \ln t}{t}$ as t-> infinity ln(t) grows slower, and t grows faster therefor we have a smaller infinity divided by a bigger infinity and so we will get 0 $\ln A= 0$ $A=1$ We can also try to do it with L'Hopitals rule: $\ln A=\lim_{x\to 0^+}\frac{\ln x}{\frac{1}{x}}$ $\ln A=\lim_{x\to 0^+}\frac{\frac{1}{x}}{ - \frac{1}{x^2}}$ $\ln A=\lim_{x\to 0^+} - \frac{x^2}{x}$ $\ln A=\lim_{x\to 0^+} - x$ $\ln A= 0$ $A= 1$ So yes you can do it with L'Hopitals rule

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