Question

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!

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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).
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Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

Mhmm right

Answer 10

not to butt in but i think your derivative is wrong

Answer 11

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

Answer 12

also the numerator is off by a minus sign

Answer 13

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(-5/x^2)\sec^2\left(\frac{5}{x}\right)}{-\frac{1}{6x^2}}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(5/x^2)\sec^2\left(\frac{5}{x}\right)}{\frac{1}{6x^2}}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\)

yes, should be this.

Answer 14

that looks better!

Answer 15

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

Answer 16

yes, now plug in that...

Answer 17

30 times (1)^2

Answer 18

How does the numerator become negative?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

Im trying to figure that out...

Answer 23

did you get everything up to and including

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) ?

Answer 24

Still thinking about the last step

Answer 25

ok take your time

Answer 26

Ahhh yeah i got it!!

Answer 27

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,

\(\large\color{slate}{\displaystyle 30\sec^2\left(\lim_{x \rightarrow ~\infty }\frac{5}{x}\right)}\)

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

so then 30 * 1 ....

Answer 32

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

Answer 33

yw