# Math Help - L'Hopital's Rule

1. ## L'Hopital's Rule

the limit of (x tan (1/x)) as x aproaches 0 from the right side

Im sorry, I dont know how to put it in LaTex Fomat :/

My first thought is to place x as $x^-1$ and take the derivative of $tan \frac {1}{x}$ but it doesn't seem right.

2. $\lim_{x\to 0^{+}}xtan(\frac{1}{x})$

This one does not appear to exist. Look at the graph and you can see.

3. Originally Posted by galactus
$\lim_{x\to 0^{+}}xtan(\frac{1}{x})$

This one does not appear to exist. Look at the graph and you can see.
in the book it says the answer is 1...but i don't see how they approached the problem...

4. Originally Posted by >_<SHY_GUY>_<
in the book it says the answer is 1...but i don't see how they approached the problem...
wait, I Am Very Sorry, it is the limit as it approaches infinity

5. Here is a graph of it. I do not know where they got 1. It looks undefined to me.

6. Originally Posted by galactus
Here is a graph of it. I do not know where they got 1. It looks undefined to me.
its an error, it aproaches infinity, does that change the problem?

7. Originally Posted by >_<SHY_GUY>_<
wait, I Am Very Sorry, it is the limit as it approaches infinity
OK. That's different then.

$\lim_{x\to {\infty}}xtan(\frac{1}{x})$

L'Hopital:

$\lim_{x\to \infty}\left(1+tan^{2}(\frac{1}{x})\right)$

$=1+\left(\lim_{x\to {\infty}}tan(\frac{1}{x})\right)^{2}$

$=1+tan^{2}\left(\frac{1}{\lim_{x\to {\infty}}}\right)=1$

8. Originally Posted by >_<SHY_GUY>_<
wait, I Am Very Sorry, it is the limit as it approaches infinity
$\lim_{x \to \infty} \frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}}$

$\lim_{x \to \infty} \frac{\sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}$

$\lim_{x \to \infty} \sec^2\left(\frac{1}{x}\right) = 1$

9. Originally Posted by galactus
OK. That's different then.

$\lim_{x\to {\infty}}xtan(\frac{1}{x})$

L'Hopital:

$\lim_{x\to \infty}\left(1+tan^{2}(\frac{1}{x})\right)$

$=1+\left(\lim_{x\to {\infty}}tan(\frac{1}{x})\right)^{2}$

$=1+tan^{2}\left(\frac{1}{\lim_{x\to {\infty}}}\right)=1$
How did you tan^2? I got lost after that :/

Originally Posted by skeeter
$\lim_{x \to \infty} \frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}}$

$\lim_{x \to \infty} \frac{\sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}$

$\lim_{x \to \infty} \sec^2\left(\frac{1}{x}\right) = 1$
I was doing just like how you were doing it, but i got stuck in the last part: sec^2 (1/x)

to make it clear, is (1/x) as x approaches infinity, is 0? what if it was approaches 0? [just 1/x]

10. $sec^{2}(u)=1+tan^{2}(u)$

It's the same thing only different

11. Originally Posted by galactus
$sec^{2}(u)=1+tan^{2}(u)$

It's the same thing only different
wait, then wouldn't it be 1 + 1 + tan^2? what happened to x?

12. $xtan(\frac{1}{x}) = x\cdot \frac{sin(\frac{1}{x})}{cos(\frac{1}{x})} = \frac{ \frac{sin(\frac{1}{x})}{\frac{1}{x}}}{cos(\frac{1} {x})}$

Now,

$\lim_{x \to \infty} \frac{ \frac{sin(\frac{1}{x})}{\frac{1}{x}}}{cos(\frac{1} {x})} \overbrace{=}^{u = \frac{1}{x}} \lim_{u \to 0} \frac{sin(u)}{u} \cdot \frac{1}{cos(u)} = \lim_{u \to 0}\frac{sin(u)}{u}\cdot \lim_{u \to 0}\frac{1}{cos (u)} = 1 \cdot 1 = 1$

13. Hello, >_<SHY_GUY>_<!

$\lim_{x\to\infty}x\tan\frac{1}{x}$
Let $u = \frac{1}{x}$

Then we have: . $\lim_{u\to0}\left(\frac{1}{u}\cdot\tan u\right) \;=\;\lim_{u\to0}\left(\frac{1}{u}\cdot\frac{\sin u}{\cos u} \right)\;=\;\lim_{u\to0}\left(\frac{\sin u}{u}\cdot\frac{1}{\cos u}\right) \;=\;1\cdot\tfrac{1}{1} \;=\;1$

14. Originally Posted by Soroban
Hello, >_<SHY_GUY>_<!

Let $u = \frac{1}{x}$

Then we have: . $\lim_{u\to0}\left(\frac{1}{u}\cdot\tan u\right) \;=\;\lim_{u\to0}\left(\frac{1}{u}\cdot\frac{\sin u}{\cos u} \right)\;=\;\lim_{u\to0}\left(\frac{\sin u}{u}\cdot\frac{1}{\cos u}\right) \;=\;1\cdot\tfrac{1}{1} \;=\;1$
Thank You Soroban! :]
Long Time No See

Im Not A Fan Of Substitution because i never get it right, but you made a good point