How can I find the limit of given trigonometric function? I am messed up on solving this question. What should I do first in order to get the answer ?
This is the trigonometric function
$$ \lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x} $$
 A: 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
$$\lim_{x\rightarrow0}\frac{\sin x}{x}=\lim_{x\rightarrow0}\frac{\sin(x/2)}{(x/2)}=1
$$
$$\lim_{x\rightarrow0}\sin(x/2)=0
$$
A: One solution, if allowed, uses Taylor series.
Built a round $x=0$, $$\sec(a+x)=\sec (a)+x \tan (a) \sec (a)+O\left(x^2\right)$$ $$(x+a)\sec(a+x)=a \sec (a)+x (\sec (a)+a \tan (a) \sec (a))+O\left(x^2\right)$$
A: Changing into cosines greatly eases the manipulation of terms.
$$
\begin{align} 
\lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x} 
& = \lim \limits_{x \rightarrow 0} \frac{a\sec(a+x) - a \sec(a)}{x} + \lim \limits_{x \rightarrow 0} \frac{x\sec(a+x)}{x} \\
& = A + B
\end{align}
$$
$$
\begin{align}
A & = \lim \limits_{x \rightarrow 0} \frac{a\sec(a+x) - a \sec(a)}{x} \\
& = a\lim \limits_{x \rightarrow 0} \frac{\cos(a) - \cos(a+x)}{\cos(a)\cos(a+x)x} \\
& = a\lim \limits_{x \rightarrow 0} 2\frac{\sin(a + x/2) sin(x/2)}{\cos(a)\cos(a+x)x} \\
& = a\lim \limits_{x \rightarrow 0} \; \frac{\sin(a + x/2)}{\cos(a)\cos(a+x)} \frac {\sin(x/2)}{x/2} \\
& = a\tan(a)sec(a)
\end{align}
$$
$$
B = \sec(a)
$$
