I recently came across the limit \begin{equation} \lim_{x \to 0} \frac{\sin(\tan^2 x)}{1 - \cos x} \end{equation} Now I know this limit can be evaluated using the half angle identity or by l'Hopital's rule but I recently tried a method that yielded the correct answer but I don't know if the method is always correct or if it's even mathematically valid. It goes as follows. \begin{equation} \lim_{x \to 0} \frac{\sin(\tan^2x)}{1 - \cos x} = \lim_{x \to 0} \frac{\sin(\sec^2 x - 1)}{1 - \cos x} \end{equation} at this point I realized I can't simply substitute $x = 0$. My intuition behind this was that although $\sec^2(0) = \sec(0)$, the functions grow at different rates, so when used in a ratio, the $\sec^2x$ answer would be greater than the $\sec x$ answer. So I made the substitution $\lim_{x \to 0} \sec x = \lim_{h \to 0} 1 + h$. This would change the limit to \begin{align} \lim_{h \to 0} \frac{\sin((1 + h)^2 - 1)}{1 - \frac{1}{1 + h}} &= \lim_{h \to 0} \frac{(1 + h)\sin(h^2 + 2h)}{h}\\ &= \lim_{h \to 0} \frac{(2 + h)(1 + h)\sin(h^2 + 2h)}{h^2 + 2h}\\ &= \lim_{h \to 0} (2 + h)(1 + h)\frac{\sin(h^2 + 2h)}{h^2 + 2h} = 2 \end{align} So is this method vlid, or was I just lucky and it happend to work for this case but would fail for most others? I originally had $h$ as an infinitesimal when I first tried this.

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    $\begingroup$ The argument is correct. It would have been clearer with the substitution $\sec x=1+h$. $\endgroup$ – André Nicolas Nov 21 '15 at 5:14
  • $\begingroup$ Your method is fine. As you say, there are many ways to solve the problem without L'Hospital. $\endgroup$ – Claude Leibovici Nov 21 '15 at 5:17
  • $\begingroup$ What do you precisely mean by "in a ratio, the $\sec^2 x$ answer would be greater than the $\sec x$ answer"? What are you comparing exactly? $\endgroup$ – Corellian Dec 11 '15 at 3:37
  • $\begingroup$ @Brody I am comparing the rate of growth. If I don't consider the square and treat $\sec^2 x $ like $\sec x$ the limit would evaluate to 1, while If I do consider the square the limit would evaluate to 2. $\endgroup$ – Jeevan Devaranjan Dec 11 '15 at 3:54

Your technique is correct. It is using substitution. Similar to substitution in definite integrals you need to change the numerical value of the limit (which you did correctly). Determining what substitution is best will come with time and practice. Knowing basic limits (like you knew $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$) is helpful. You should be able to find plenty more examples of this technique here on SE.

Here is an example I recently saw that you may want to think about what substitution you'd use:

Calculate a limit $\lim_{x \to \pi/2} \frac{\sqrt[4]{ \sin x} - \sqrt[3]{ \sin x}}{\cos^2x}$


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