# Thread: needing help on evalutating this limit..

1. ## needing help on evalutating this limit..

hello, sorry for using my first post on this help forum ^^; I've been trying to ask this question everywhere on the net and no one seems to be able to help. So I'm trying here. Anyway

I'm trying to find the limit of x tan (8/x) as x approaches to infinity, using the l'hospital's rule.

So far I've applied the rule, but I'm stuck at the last part. Here's my work so far.

Using the l'hospital's rule it turned out like this:
$\displaystyle sec^2 (8/x) * -8/x^2 / (-8/x^2)$

+the$\displaystyle -8/x^2$'s cancel out leaving me with
$\displaystyle sec^2 (8/x)$

so now this is where I'm stuck...When you input infinity in (8/x) , I can't seem to find where the values are heading to... for some examples

$\displaystyle 8/80 = .1$

$\displaystyle 8/90 = 0.08888~$

$\displaystyle 8/100$

so.. would these values be heading towards 0 ? For my final answer I ended up getting, sec^2 (0) = 1 ... I think I made a mistake around here somehwere, but I don't know what.

if you can help, it would be greatly appreciated!

2. Originally Posted by endiv
hello, sorry for using my first post on this help forum ^^; I've been trying to ask this question everywhere on the net and no one seems to be able to help. So I'm trying here. Anyway

I'm trying to find the limit of x tan (8/x) as x approaches to infinity, using the l'hospital's rule.

So far I've applied the rule, but I'm stuck at the last part. Here's my work so far.

Using the l'hospital's rule it turned out like this:
$\displaystyle sec^2 (8/x) * -8/x^2 / -8/x^2$

+the$\displaystyle -8/x^2$'s cancel out leaving me with
$\displaystyle sec^2 (8/x)$

so now this is where I'm stuck...When you input infinity in (8/x) , I can't seem to find where the values are heading to... for some examples

$\displaystyle 8/80 = .1$

[tex]8/90 = 0.08888~

$\displaystyle 8/100$

so.. would these values be heading towards 0 ? For my final answer I ended up getting, sec^2 (0) = 1 ... I think I made a mistake around here somehwere, but I don't know what.

if you can help, it would be greatly appreciated!
If you have to apply the L'hopital's rule make it easier on yourself

$\displaystyle \lim_{n\to\infty}\frac{\tan\left(\frac{1}{n}\right )}{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{\cos\lef t(\frac{1}{n}\right)}\cdot\lim_{n\to\infty}\frac{\ sin\left(\frac{1}{n}\right)}{\frac{1}{n}}=\lim_{n\ to\infty}\frac{\sin\left(\frac{1}{n}\right)}{\frac {1}{n}}$

Differentiating top and bottom givse

$\displaystyle \lim_{n\to\infty}\frac{\cos\left(\frac{1}{n}\right )\cdot\frac{-1}{n^2}}{\frac{-1}{n^2}}=\lim_{n\to\infty}\cos\left(\frac{1}{n}\ri ght)=1$

3. Ah, I found it ! I was confused comparing sin 8/x to the sin 1/x in your example. I thought that plugging the infinite in the 8/x would give me the limit so I ended up finding all this wierd answers. Thanks again!