# Math Help - L'Hospital's Rule

1. ## L'Hospital's Rule

Find the limit of x(tan(8/x)) as it approaches infinity.

If I plug in infinity, I get (infinity * 0) so the problem is an indeterminate form.

So I find the limit of tan(8/x) / (1/x). This is (0 * 0) so I can use l'Hospital's rule:

sec^2(8/x)(-8x^-2) / 1

That still is (0 * 0), so I can use l'Hospital's again. I get:

(16x^-3) (2sec(8/x)(-8x^-2)

At this time I realize that this doesn't look like it's helping me, so I probably did something wrong. Can someone look over my work, tell me if I'm on the right track, and if not, tell me what I did wrong? Thanks.

2. Originally Posted by Maziana
Find the limit of x(tan(8/x)) as it approaches infinity.

If I plug in infinity, I get (infinity * 0) so the problem is an indeterminate form.

So I find the limit of tan(8/x) / (1/x). This is (0 * 0) so I can use l'Hospital's rule:

sec^2(8/x)(-8x^-2) / 1

That still is (0 * 0), so I can use l'Hospital's again. I get:

(16x^-3) (2sec(8/x)(-8x^-2)

At this time I realize that this doesn't look like it's helping me, so I probably did something wrong. Can someone look over my work, tell me if I'm on the right track, and if not, tell me what I did wrong? Thanks.
Your derivative of $\tan \left( \frac{8}{x}\right)$ is wrong. You need to use the chain rule.

3. Originally Posted by mr fantastic
Your derivative of $\tan \left( \frac{8}{x}\right)$ is wrong. You need to use the chain rule.
Actually the derivative of $\tan{\left(\frac{8}{x}\right)}$, which is $\sec^2{\left(\frac{8}{x}\right)}[-8x^{-2}]$ is correct.

It is the derivative of $\frac{1}{x}$ in the denominator that is incorrect.

4. That's right; thanks. 1/x should go to -1/x^2. So it should be:

sec^2(8/x)(-8x^-2) / -1/x^2

That's still (0 * 0), so I can use l'Hospital's again. But I still don't see how that helps; it just makes the problem even more complicated.

5. Try setting $t=1/x$ now when $x\to\infty$, $t\to0$

Then you can calculate the limit:

$\lim\limits_{t\to0}\frac{\tan(8t)}{t}$.