L'Hospital's Rule

[elcatracho]

Hello I had two questions about problems that use L'Hospital's rule. The instructions are find the indicated limit. Use L'Hospital's rule IF necesary. the first question is lim of x^x as x gets closer to 0+. So far I have done x times lnx. I dont know what to do from here. The next problem is lim of Sin(x)^(tan(x)) as x approaches pi/2-. Again, so far is did tan(x) times Sin(x) but i dont know what to do from there. When I did it on the calculator i got 1 for both answers but I'd really like to know how to do it by hand. Thank You

 

[pka]

\(\displaystyle \L
\begin{array}{l}
y = x^x \quad \Rightarrow \quad \ln (y) = x\ln (x) \\
\lim _{x \to 0^ + } \ln (y) = \lim _{x \to 0^ + } \frac{{\ln (x)}}{{\frac{1}{x}}} = \limits^H \lim _{x \to 0^ + } \frac{{\frac{1}{x}}}{{\frac{{ - 1}}{{x^2 }}}} = 0 \\
\lim _{x \to 0^ + } x^x = \lim _{x \to 0^ + } e^{\ln (y)} = e^{\lim _{x \to 0^ + } \ln (y)} = 1 \\
\end{array}\)

 

[soroban]

Hello, elcatracho!

These are "logarithmic limits", very tricky until you get used to them (if ever).
Here's the second one . . .

\(\displaystyle \L\lim_{x\to\frac{\pi}{2}} (\sin x)^{\tan x}\)

\(\displaystyle \text{Let }\L\,y\;=\;(\sin x)^{\tan x}\)

\(\displaystyle \text{Take logs: }\L\,\ln y \;= \;\ln(\sin x)^{\tan x} \;= \;\tan x\cdot\ln(\sin x)\)

\(\displaystyle \text{If }x\to\frac{\pi}{2},\text{ the expression goes to: }\L\tan\frac{\pi}{2}\cdot\ln\left(\sin\frac{\pi}{2}\right) \:=\\infty)(0)\)

\(\displaystyle \text{Re-write it: }\L\,\ln y \;= \;\frac{\ln(\sin x)}{\cot x}\)\(\displaystyle \;\text{ . . . which goes to }\frac{0}{0} \text{ . . . good!}\)

\(\displaystyle \text{Apply L'Hopital: }\L\,\ln y \;= \;\frac{\frac{\cos x}{\sin x}}{-\csc^2x} \;= \;-\sin x\cdot\cos x\)

\(\displaystyle \text{Take the limit: }\L\,\lim_{x\to\frac{\pi}{2}} \ln y \;= \;\lim_{x\to\frac{\pi}{2}}[-\sin x\cdot\cos x] \;= \;-(1)(0) \;= \;0\)

\(\displaystyle \text{Since }\L\,\lim_{x\to\frac{\pi}{2}} \ln y \;= \;0\text{, then: }\L\,y\:=\:e^0\:=\:1\)