- #1
kdinser
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Today we started on infinite series, I'm getting the material just fine and able to do most of the problems, but one is giving me problems.
[tex]\lim_{n\to{a}} 2n/\sqrt{n^2+1}[/tex]
I recognized that [tex]\infty/\infty[/tex] so I can use L'Hopital's rule. So taking the derivative of the numerator and denominator I get.
[tex]\frac{d}{dn} 2n = 2[/tex]
and
[tex]\frac{d}{dn} \sqrt{n^2+1} = \frac{n}{\sqrt{n^2+1}}[/tex]
Somehow the solutions manual is getting
[tex]2/\sqrt{1+(1/n^2)}[/tex]
and a final answer of 2.
I can't see how they turned what I get for the derivative in the denominator into what they use. L'Hopital's rule twice would get rid of the n on top and put a 2 there, but that wouldn't change the square root.
thanks for any help
[tex]\lim_{n\to{a}} 2n/\sqrt{n^2+1}[/tex]
I recognized that [tex]\infty/\infty[/tex] so I can use L'Hopital's rule. So taking the derivative of the numerator and denominator I get.
[tex]\frac{d}{dn} 2n = 2[/tex]
and
[tex]\frac{d}{dn} \sqrt{n^2+1} = \frac{n}{\sqrt{n^2+1}}[/tex]
Somehow the solutions manual is getting
[tex]2/\sqrt{1+(1/n^2)}[/tex]
and a final answer of 2.
I can't see how they turned what I get for the derivative in the denominator into what they use. L'Hopital's rule twice would get rid of the n on top and put a 2 there, but that wouldn't change the square root.
thanks for any help
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