2

Feeling like i did this wrong

$\displaystyle \lim_{x \to 0}\frac{\tan x-x}{x^3}$ $\to$ $\displaystyle \lim_{x \to 0}\frac{\sec^2x-1}{3x^2}$

$\displaystyle \lim_{x \to 0}\frac{2\tan x\sec^2x}{4x}$ $\to$ $\displaystyle \lim_{x \to 0}\frac{2\sec^2x\sec^2x+2\tan x(2\tan x\sec^2x)}{6}$

Simplified

$\displaystyle \lim_{x \to 0}\frac{\sec^2x(\sec^2x+4\tan^2x)}{3}$

Not really sure what to do at this point

iostream007
  • 4,529
John Snow
  • 217
  • 3
  • 17
  • 5
    Why not plug in zero? It looks like you're done to me beyond that. – Alex Wertheim May 10 '13 at 05:01
  • Okay, I was just hoping for a confirmation i did this right.Thanks!

    For the limit i got 1/2

     – John Snow May 10 '13 at 05:09
  • It should be $3x^2$ and not $2x^2$ right? –  May 10 '13 at 05:13
  • Yes good catch! The limit would then be 1/3 – John Snow May 10 '13 at 05:37
  • 1
    Don't use $\to$ to mean $=$. – Andrés E. Caicedo May 10 '13 at 15:35
  • 1
    We encourage users to accept answers they find helpful: you can accept exactly one answer per question asked. Just click on the $\Large \checkmark$ to the left of the answer you'd like to accept. (And...you get two reputation points for each question which you accept an answer for!) You can also upvote as many answers as you'd like (just click on the $\uparrow$ above the "vote count" to the left of the answer you'd like to upvote.) – amWhy May 15 '13 at 23:58

4 Answers4

5

Hint: The function you've reached is defined and continuous at $x=0$.

Cameron Buie
  • 102,994
2

I think the easier way to find this limit is use series expansion of $\tan x$ if you know this.

$$\tan x =x + \frac{1}{3}x^3+\frac{2}{15}x^5+\cdots$$

Argha
  • 4,671
2

You could also note that

$$ \lim_{x \to 0} \frac{2 \tan x \sec^2 x}{6x} = \lim_{x \to 0}\frac{2 \sec^2 x}{6} $$ since $$ \lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot 1 = 1. $$

JavaMan
  • 13,153
0

The Pythagorean Identity is of help here: starting from your first application of l'Hopital's Rule, you could then write

$$\lim_{x \rightarrow 0} \ \frac{\sec^2 x \ - \ 1}{3 x^2} \ ^{*} \ = \ \lim_{x \rightarrow 0} \ \frac{(\tan^2 x \ + \ 1 ) \ - \ 1}{3 x^2} \ = \ \lim_{x \rightarrow 0} \ \frac{\tan^2 x }{3 x^2} $$

  • there is an error in your differentiation of $x^3$

$$= \ \lim_{x \rightarrow 0} \ \frac{\sin^2 x}{3 x^2 \ \cdot \ \cos^2 x } \ = \ \lim_{x \rightarrow 0} \ \frac{1}{3} \ \cdot \ (\frac{\sin x}{x})^2 \ \cdot \ \frac{1}{\cos^2 x} \ , $$

with the limit of the middle factor having a familiar value.

colormegone
  • 10,842