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

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 $\sec^2x$?

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  • $\begingroup$ Why do you stop understanding L'Hopital halfway through? $\endgroup$ – J.G. Apr 8 '18 at 14:53
  • $\begingroup$ I see where I made an error. I did not see $sec^2\ (0)=1$. $\endgroup$ – Jinzu Apr 8 '18 at 14:56
  • $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Apr 8 '18 at 14:58
  • $\begingroup$ 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 $\tan{x}$. $\endgroup$ – Netravat Pendsey Apr 8 '18 at 15:34
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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 $$ \lim_{x\to0} \sec^2 x = 1. $$

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

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They have used L'Hopital's Rule.

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