Question

Find the Limit
Lim (9/x)[ (1/arctanx) - (1/x)]
X->0

Answer 1

9/x is indeterminate as x __> 0 so I guess you need to use L'hopitals rule

Answer 2

i'm not sure though

Answer 3

yes i know .. but what is the excat limit?

Answer 4

Let \(x=\tan y\), then as \(x\to0\) you have that \(y\to0\).
\[\begin{align*}\lim_{x\to0}\frac{9}{x}\left(\frac{1}{\arctan x}-\frac{1}{x}\right)&=\lim_{y\to0}\frac{9}{\tan y}\left(\frac{1}{y}-\frac{1}{\tan y}\right)\\\\
&=9\lim_{y\to0}\frac{\tan y-y}{y\tan^2 y}\\\\
&=9\lim_{y\to0}\frac{\tan y-y}{y\tan^2y}\end{align*}\]
Evaluating directly gives the indeterminate \(\dfrac{0}{0}\), so let's try L'Hopital's rule:
\[\begin{align*}9\lim_{y\to0}\frac{\sec^2y-1}{\tan^2y+2y\tan y\sec^2y}&=9\lim_{y\to0}\frac{\tan^2y}{\tan^2y+2y\tan y\sec^2y}\\\\
&=9\lim_{y\to0}\frac{\tan y}{\tan y+2y\sec^2y}\\\\
&=9\lim_{y\to0}\frac{\sin y\cos y}{\sin y\cos y+2y}\\\\
&=9\lim_{y\to0}\frac{\frac{1}{2}\sin 2y}{\frac{1}{2}\sin2y+2y}\\\\
&=9\lim_{y\to0}\frac{\sin 2y}{\sin2y+4y}
\end{align*}\]
Evaluating directly gives \(\dfrac{0}{0}\), so let's use L'Hopital's again.
\[9\lim_{y\to0}\frac{2\cos 2y}{2\cos2y+4}=9\times\frac{2}{2+4}=3\]