Question

check my work

Answer 1

Find:
\[\lim_{x \rightarrow 0^{+}}[-\frac{1}{\ln x}]^x\]
consider
\[y=(-\frac{1}{\ln x})^x\]\[\ln y=x.\ln(-\frac{1}{\ln x})\]
\[\lim_{x \rightarrow 0^+}\ln y=\lim_{x \rightarrow 0^+}\frac{\ln(-\frac{1}{\ln x})}{\frac{1}{x}} \space \space \space (\frac{\infty}{\infty})\]
\[\lim_{x \rightarrow 0^+}\ln y=\lim_{x \rightarrow 0^+}-\ln x . \frac{1}{(\ln x)^2}.\frac{1}{x}.\frac{1}{-\frac{1}{x^2}}\]\[\lim_{x \rightarrow 0^+} \ln y=\lim_{x \rightarrow 0^+} \frac{x}{\ln x}\]\[\lim_{x \rightarrow 0^+}\ln y=\lim_{x \rightarrow 0^+}\frac{x}{\frac{1}{\ln x}} \space \space \space \space \space \space (\frac{0}{0})\]
\[\lim_{x \rightarrow 0^+}\ln y=\lim_{x \rightarrow 0^+}\frac{1}{-\frac{1}{(\ln x)^2}.\frac{1}{x}}\]
\[\lim_{x \rightarrow 0^+}\ln y=-\lim_{x \rightarrow 0^+} \frac{(\ln x)^2}{\frac{1}{x}} \space \space \space \space \space (\frac{\infty}{\infty})\]
\[\lim_{x \rightarrow 0+}\ln y=-\lim_{x \rightarrow 0^+}\frac{2\ln x. \frac{1}{x}}{-\frac{1}{x^2}}\]\[\lim_{x \rightarrow 0^+}\ln y=2\lim_{x \rightarrow 0^+}x \ln x =2 \lim_{x \rightarrow 0^+}\frac{\ln x}{\frac{1}{x}} \space \space \space \space \space (\frac{\infty}{\infty})\]\[\lim_{x \rightarrow 0^+}\ln y=2\lim_{x \rightarrow 0+}\frac{1}{x}.\frac{1}{-\frac{1}{x^2}}=-2\lim_{x \rightarrow 0^+}\frac{x^2}{x}=0\]
\[\lim_{x \rightarrow 0^+}\ln y=0 \space \implies \lim_{x \rightarrow 0^+}y=1\]
\[\lim_{x \rightarrow 0^+} [-\frac{1}{\ln x}]^x=1\]

Answer 2

i think theres a mistake.......

Answer 3

cut everything after
\[\lim_{x \rightarrow 0^+}\ln y=\lim_{x \rightarrow 0^+}\frac{x}{\ln x}\]
I don't know how I made it into
\[\frac{x}{\frac{1}{\ln x}}\]
for some reason

Answer 4

I can't understand how to proceed after
\[\lim_{x \rightarrow 0^+}\ln y=\lim_{x \rightarrow 0^+}\frac{x}{\ln x}\]

Answer 5

may be let \(x = 1/y\) :
\[\lim\limits_{x\to 0^+}\dfrac{x}{\ln x} = \lim\limits_{y\to\infty} \dfrac{1/y}{\ln(1/y)} = 0\]

Answer 6

Ah!! thanks a ton....

Answer 7

That would work

Answer 8

of course I'll use a different variable, im already using y

Answer 9

I think
That makes it 0/infinity form

Answer 10

any other method?

Answer 11

\[\lim\limits_{x\to 0^+}\dfrac{x}{\ln x} = \lim\limits_{y\to\infty} \dfrac{1/y}{\ln(1/y)} =- \lim\limits_{y\to\infty} \dfrac{1}{y\ln(y)}= 0\]

Answer 12

Ok, I understand it now, it tends to -1/infinity or 0, that was a clever trick, I'll definitely remember it!!

Answer 13

\[\lim_{x \rightarrow 0}\frac{x^2e^x}{\sin^2(3x)}=\lim_{x \rightarrow 0}\frac{x^2}{\sin^2(3x)}.\lim_{x \rightarrow 0}e^x=\lim_{x \rightarrow 0}\frac{x^2}{\sin^2(3x)}\]
0/0 form so we apply L'Hopital's rule
\[\lim_{x \rightarrow 0} \frac{2x}{3\sin(6x)}=\frac{2}{3}\lim_{x \rightarrow 0}\frac{x}{\sin(6x)}\]
Now as
\[x \rightarrow 0 \space \space \space \implies 6x \rightarrow 0\]
and \[\lim_{x \rightarrow 0}\frac{\sin x}{x}=\lim_{x \rightarrow 0}\frac{x}{\sin x}=1\]
\[\therefore \frac{2}{3}.\frac{1}{6}\lim_{6x \rightarrow 0 }\frac{6x}{\sin(6x)}=\frac{1}{9}.1=\frac{1}{9}\]

@ganeshie8

Answer 14

Looks good to me !