## anonymous one year ago I'm trying to evaluate this limit using L'Hopital's rule... But apparently I'm doing something wrong! Would appreciate if someone could spot what it is I'm doing wrong!

1. anonymous

$\lim _{x \rightarrow \infty} 6xtan(5/x)$

2. anonymous

So I rewrote it as$\lim _{x \rightarrow \infty} \frac{ \tan(5/x) }{ \frac{ 1 }{ 6x } }$

3. anonymous

Once applying L'Hopital's rule I get... $\frac{ (\frac{ 5 }{ x^2 })\sec^2(5/x) }{ -36x }$

4. anonymous

And I believe that the limit of the above equation would be equal to 0... Why is that wrong?

5. SolomonZelman

can you use the numerical approach. I would do so here.... I am not coming up with a way to re-write this so that you can differentiate top and bottom.... with numerical approach it seems to approach 30 (I plugged in a couple of values). -------------------------------------------------------

6. anonymous

Hmmm... so just to plug in large numbers?

7. SolomonZelman

well, I did it for the check. Just that we know the correct answer. I plugged very large numbers to see how the function behaves the more x tends towards infinity.

8. SolomonZelman

I am trying to think f a good way to this one, if there is.

9. anonymous

Mhmm right

10. anonymous

not to butt in but i think your derivative is wrong

11. anonymous

$\frac{d}{dx}[\frac{1}{6x}]=-\frac{1}{6x^2}$

12. anonymous

also the numerator is off by a minus sign

13. SolomonZelman

$$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(-5/x^2)\sec^2\left(\frac{5}{x}\right)}{-\frac{1}{6x^2}}}$$ $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(5/x^2)\sec^2\left(\frac{5}{x}\right)}{\frac{1}{6x^2}}}$$ $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}$$ yes, should be this.

14. anonymous

that looks better!

15. SolomonZelman

in my third line, dividing by 1/6 is multiplying times 6. and 1/x^2 cancel

16. SolomonZelman

yes, now plug in that...

17. SolomonZelman

30 times (1)^2

18. anonymous

How does the numerator become negative?

19. SolomonZelman

because you are taking the derivative of 5/x and that is a negative exponent

20. anonymous

Oh oh oh!!! Yeah just a confusion sorry!

21. SolomonZelman

derivat. of (5/x) is the chain rule for your angle....do you get why it is 30 ?

22. anonymous

Im trying to figure that out...

23. SolomonZelman

did you get everything up to and including $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}$$ ?

24. anonymous

Still thinking about the last step

25. SolomonZelman

26. anonymous

Ahhh yeah i got it!!

27. SolomonZelman

yes, so then we have: $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}$$ applying our limit properties, $$\large\color{slate}{\displaystyle 30\sec^2\left(\lim_{x \rightarrow ~\infty }\frac{5}{x}\right)}$$

28. SolomonZelman

$$\large\color{slate}{\displaystyle 30\sec^2\left(0\right)}$$

29. anonymous

Thanks so much for the help @SolomonZelman and @satellite73 :)

30. SolomonZelman

yes, and then you know sec^2(0)=1/cos^2(0)=1/1^2=1

31. SolomonZelman

so then 30 * 1 ....

32. anonymous

Yes, I understand the rest :) Thanks so much!

33. SolomonZelman

yw