limit x->inf using L'H rule, tan^-1(x)/((x^-1)-5)

[Joseph007]

Find the limit x->inf of tan^-1(x)/((x^-1)-5)

I use L'H's rule, getting (1+x^2)^-1/(-(x^2))^-1
and again to get -2x/2x, to get -1, but it is wrong, what is going on?[/tex]

 

[pka]

This is the problem as you have written it: \(\displaystyle \L
\frac{{\arctan (x)}}{{\frac{1}{{x}} - 5}}\).

If that is correct, then one would not use l’Hopital’s rule.
The limit is \(\displaystyle \frac{\pi }{{10}}\).

If this is NOT the problem as you have written it, please correct it it.

 

[Joseph007]

that is it, I found someone with a ti89, but I dont know how it gets that

 

[galactus]

\(\displaystyle tan^{-1}({\infty})=\frac{{\pi}}{2}\)

What does the denominator converge to as x approaches infinity?.

 

[pka]

\(\displaystyle tan^{-1}({\infty})=\frac{{\pi}}{2}\)

PLEASE, PLEASE, PLEASE NEVER write \(\displaystyle tan^{-1}({\infty})=\frac{{\pi}}{2}\).
\(\displaystyle \infty\) is NOT A NUMBER.
\(\displaystyle \lim\limits_{x \to - \infty } \;\arctan (x) = \frac{{ - \pi }}{2}\).

 

[galactus]

Sorry.