What is the limit as x approaches 0 of $\frac{cot x}{ln x}$ $cotx/lnx$ ?

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Answer 1 · 2014-12-11T07:02:46

Since $ln x$ $ln x$ is only defined when $x > 0$ $x>0$ , we can only talk about the right-hand limit:

$lim_{x to 0^+}{cot x}/{ln x}$

Since

${(lim_{x to 0^+}cot x=lim_{x to 0^+}{cos x}/{sin x}={1}/{0^+}=+infty),(lim_{x to 0^+}ln x=-\infty):}$ ,

by applying l'H $hat{"o"}$ pital's Rule ( $infty$ / $infty$ ),

$=lim_{x to 0^+}{-csc^2x}/{1/x}$

by $csc x=1/{sin x}$ ,

$=-lim_{x to 0^+}x/{sin^2x}$

by l'H $hat{"o"}$ pital's Rule ( $0"/"0$ ),

$=-lim_{x to 0^+}1/{2sinx cosx}=-1/0^+=-infty$

The graph of $y={cot x}/{ln x}$ looks like:

enter image source here

As shown above, as $x$ approaches 0 from the right, the graph is shooting down toward $-infty$ .

What is the limit as x approaches 0 of cotx/lnxcotxlnxcotx/lnx?