# Find $\lim\limits_{x\to 0}\frac{\tan x - x}{x^3}$

$$\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\sec^2 x-1}{3x^2}=\lim_{x\to0}\frac{2\sec^2x\tan x}{6x}\\ =\frac{1}{3}\lim_{x\to0}\frac{\tan x}{x}=\frac{1}{3}\lim\limits_{x\to0}\frac{\sec^2x}{1}=\frac{1}{3}$$

My question is about the last line. Why do they set the two expressions $\frac{\tan x}{x}$ and $\frac{\sec^2x}{1}$ equal to each other and why is the one-third in both sets of expressions, instead of multiplying the two expressions? Also what happens to the $x$ under the $\sec^2x$?

• Why do you stop understanding L'Hopital halfway through? – J.G. Apr 8 '18 at 14:53
• I see where I made an error. I did not see $sec^2\ (0)=1$. – Jinzu Apr 8 '18 at 14:56
• MathJax tip: common functions such as trig functions can be written using the format \functionName (e.g. \tan and \sec) which automatically applies proper spacing before and after it. Also, use the format $$...$$ for display style formatting so you don't need to write out \Large or \lim\limits. See also edits. – Simply Beautiful Art Apr 8 '18 at 14:58
• I want to point out one thing,(not related to your doubt or method) that the solution is extremely short if you use expansion on $\tan{x}$. – Netravat Pendsey Apr 8 '18 at 15:34

## 3 Answers

It doesn't say $\dfrac{\tan x} x$ is equal to $\dfrac{\sec^2 x} 1;$ rather it says $\lim\limits_{x\to0} \dfrac{\tan x} x = \lim\limits_{x\to0} \dfrac{\sec^2 x} 1.$

Their limits as $x\to0$ are equal; the functions themselves are not.

The reason for the conclusion that they are equal is the same as the reason for the equalities in $$\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\sec^2 x-1}{3x^2} =\lim_{x\to0}\frac{2\sec^2x\tan x}{6x}.$$ That reason is L'Hopital's rule.

The next equality after that is not deduced from L'Hopital's rule, but from the equality $$\lim_{x\to0} \sec^2 x = 1.$$

This is a matter of applying L'Hopital's Rule, which is just plain silly. Of course that$$\lim_{x\to0}\frac{\tan x}x=\tan'(0)=1.$$

They have used L'Hopital's Rule.