Limit without Lhospital rules

Question

please how can i solve this limit ((1/cos x)-(1/cos a))/(x-a) where x aproches a. Thank you for you help. Unfortunately i dont understand the way how to solve limits of this type.

$=\frac{\sin((x-a)/2)}{(x-a)/2}\frac{\sin((x+a)/2)}{\cos(x)\cos(a)}$. Note that $\frac{\sin((x-a)/2)}{(x-a)/2}\to 1$, $\sin((x+a)/2)\to\sin(a)$ and $\cos(x)\to\cos(a)$. Therefore, the limit is $\frac{\sin(a)}{\cos^2(a)}$. — conditionalMethod, Dec 05 '19 at 08:52
You will get answer in 1 step directly if you use L hospitals rule — user600016, Dec 05 '19 at 08:56
@user600016 But how is that useful when the title explicitly states without l'Hopital's rule...? — StackTD, Dec 05 '19 at 08:59

StackTD · Answer 1 · 2019-12-05T09:00:24.367

Note that $$\lim_{x \to a}\frac{\frac{1}{\cos x}-\frac{1}{\cos a}}{x-a}$$ is of the form $$\lim_{x \to a}\frac{f(x)-f(a)}{x-a}$$ with $f(x)=\tfrac{1}{\cos x}$, which is exactly the definition of $f'(a)$, if the limit exists.

Of course, you might encounter this limit because you're trying to find the derivative of $f(x)=\tfrac{1}{\cos x}$ from the limit definition. In that case, you can proceed as follows to avoid circular reasoning: $$\begin{align}\lim_{x \to a}\frac{\frac{1}{\cos x}-\frac{1}{\cos a}}{x-a} & = \lim_{x \to a}\frac{\cos a - \cos x}{\left(x-a\right)\cos x\cos a} \\[6pt] & = \lim_{x \to a}\frac{2\sin\frac{x-a}{2}\sin\frac{x+a}{2}}{\left(x-a\right)\cos x\cos a} \tag{Simpson's formula}\\[6pt] & = \lim_{x \to a}\frac{\sin\frac{x-a}{2}}{\frac{x-a}{2}}\frac{\sin\frac{x+a}{2}}{\cos x\cos a} \\[6pt] & = \underbrace{\lim_{x \to a}\frac{\sin\frac{x-a}{2}}{\frac{x-a}{2}}}_{\to 1}\lim_{x \to a}\frac{\sin\frac{x+a}{2}}{\cos x\cos a} \\[6pt] & = \tan a \sec a \end{align}$$

score 0 · Answer 2 · answered Dec 05 '19 at 08:29

0

By the definition of a derivative, this limit is $\sec^\prime a=\sec a\tan a$. (This may be more obvious if you define $h:=x-a$.)

answered Dec 05 '19 at 08:29

J.G.

115,835

Limit without Lhospital rules

2 Answers2