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I am messed up on solving this question. What should I do first in order to get the answer ?

This is the trigonometric function

$$ \lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x} $$

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  • $\begingroup$ Use l'hopital's rule. $\endgroup$ – albo May 21 '16 at 7:23
  • $\begingroup$ L-hospital is not allowed as per the question. $\endgroup$ – Saugat Pokharel May 21 '16 at 7:24
  • $\begingroup$ Then use $\cos(a+x)=\cos a\cos x-\sin a\sin x$, approximate $\sin x$ as $x$ and $\cos x$ as 1. Or, as charMD says, use the fact that your expression is just the derivative of $t\sec t$ at $t=a$. $\endgroup$ – almagest May 21 '16 at 7:31
  • $\begingroup$ @almagest The question is given in terms of $\sec$, not $\cos$ or $\sin$. $\endgroup$ – Mark May 21 '16 at 7:32
  • $\begingroup$ A simple way (new view): Let $f(a)=a\sec a$. Then it is just equal to $f'(a)$ by the definition. $\endgroup$ – Paul May 21 '16 at 8:55
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Method 1: This is the definition of $$\frac d{da}(a\sec a)=\sec a+a\sec a\tan a$$ Method 2: Change those secants to cosines $$ \begin{align} \lim_{x\rightarrow0}\frac{(a+x)\cos a-a\cos(a+x)}{x\cos a\cos(a+x)}&=\lim_{x\rightarrow0}\frac{a\cos a+x\cos a-a\cos a\cos x+a\sin a\sin x}{x\cos a\cos(a+x)}\\ &=\lim_{x\rightarrow0}\frac{a\cos a(1-\cos x)+x\cos a+a\sin a\sin x}{x\cos a\cos(a+x)}\\ &=\lim_{x\rightarrow0}\left[\frac{2a\sin^2(x/2)}{x\cos(a+x)}+\frac1{\cos(a+x)}+\frac{a\sin a\sin x}{x\cos a\cos(a+x)}\right]\\ &=0+\sec a+a\tan a\sec a \end{align} $$ In the last line we have used the limits $$\lim_{x\rightarrow0}\frac{\sin x}{x}=\lim_{x\rightarrow0}\frac{\sin(x/2)}{(x/2)}=1 $$ $$\lim_{x\rightarrow0}\sin(x/2)=0 $$

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One solution, if allowed, uses Taylor series.

Built a round $x=0$, $$\sec(a+x)=\sec (a)+x \tan (a) \sec (a)+O\left(x^2\right)$$ $$(x+a)\sec(a+x)=a \sec (a)+x (\sec (a)+a \tan (a) \sec (a))+O\left(x^2\right)$$

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Changing into cosines greatly eases the manipulation of terms. $$ \begin{align} \lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x} & = \lim \limits_{x \rightarrow 0} \frac{a\sec(a+x) - a \sec(a)}{x} + \lim \limits_{x \rightarrow 0} \frac{x\sec(a+x)}{x} \\ & = A + B \end{align} $$ $$ \begin{align} A & = \lim \limits_{x \rightarrow 0} \frac{a\sec(a+x) - a \sec(a)}{x} \\ & = a\lim \limits_{x \rightarrow 0} \frac{\cos(a) - \cos(a+x)}{\cos(a)\cos(a+x)x} \\ & = a\lim \limits_{x \rightarrow 0} 2\frac{\sin(a + x/2) sin(x/2)}{\cos(a)\cos(a+x)x} \\ & = a\lim \limits_{x \rightarrow 0} \; \frac{\sin(a + x/2)}{\cos(a)\cos(a+x)} \frac {\sin(x/2)}{x/2} \\ & = a\tan(a)sec(a) \end{align} $$ $$ B = \sec(a) $$

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