find the limit x-->0+ of f(x) = sin(x) ln(x) thanks
M mathhelp New member Joined Nov 12, 2006 Messages 25 Dec 17, 2006 #1 find the limit x-->0+ of f(x) = sin(x) ln(x) thanks
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Dec 17, 2006 #2 Re: another limit Hello, mathhelp! I used L'Hopital twice . . . Find: \(\displaystyle \,\lim_{x\to0^+}\sin x\cdot\ln x\) Click to expand... The function goes to: \(\displaystyle \,0(-\infty)\) Rewrite as: \(\displaystyle \:\lim_{x\to0^+}\L\frac{\ln x}{\csc x}\) . . . which goes to: \(\displaystyle \,\frac{-\infty}{\infty}\) Apply L'Hopital: \(\displaystyle \L\,\frac{\frac{1}{x}}{-\csc x\cdot\cot x} \:=\:-\frac{\sin x\cdot\tan x}{x}\) . . . which goes to \(\displaystyle \frac{0}{0}\) Apply L'Hopital again: \(\displaystyle \L\,-\frac{\sin x\cdot\sec^2x\,+\,\cos x\cdot\tan x}{1}\) Therefore: \(\displaystyle \,\lim_{x\to0^+}\left(-\sin x\cdot\sec^2x\,-\,\cos x\cdot\tan x\right) \:=\:-0\cdot1^2\,-\,1\cdot0 \:=\:0\)
Re: another limit Hello, mathhelp! I used L'Hopital twice . . . Find: \(\displaystyle \,\lim_{x\to0^+}\sin x\cdot\ln x\) Click to expand... The function goes to: \(\displaystyle \,0(-\infty)\) Rewrite as: \(\displaystyle \:\lim_{x\to0^+}\L\frac{\ln x}{\csc x}\) . . . which goes to: \(\displaystyle \,\frac{-\infty}{\infty}\) Apply L'Hopital: \(\displaystyle \L\,\frac{\frac{1}{x}}{-\csc x\cdot\cot x} \:=\:-\frac{\sin x\cdot\tan x}{x}\) . . . which goes to \(\displaystyle \frac{0}{0}\) Apply L'Hopital again: \(\displaystyle \L\,-\frac{\sin x\cdot\sec^2x\,+\,\cos x\cdot\tan x}{1}\) Therefore: \(\displaystyle \,\lim_{x\to0^+}\left(-\sin x\cdot\sec^2x\,-\,\cos x\cdot\tan x\right) \:=\:-0\cdot1^2\,-\,1\cdot0 \:=\:0\)
