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.

Answer 1

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

Answer 2

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.

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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: $$\begin{align}\lim_{x \to a}\frac{\frac{1}{\cos x}-\frac{1}{\cos a}}{x-a} & = \lim_{x \to a}\frac{\cos a - \cos x}{\left(x-a\right)\cos x\cos a} \\[6pt] & = \lim_{x \to a}\frac{2\sin\frac{x-a}{2}\sin\frac{x+a}{2}}{\left(x-a\right)\cos x\cos a} \tag{Simpson's formula}\\[6pt] & = \lim_{x \to a}\frac{\sin\frac{x-a}{2}}{\frac{x-a}{2}}\frac{\sin\frac{x+a}{2}}{\cos x\cos a} \\[6pt] & = \underbrace{\lim_{x \to a}\frac{\sin\frac{x-a}{2}}{\frac{x-a}{2}}}_{\to 1}\lim_{x \to a}\frac{\sin\frac{x+a}{2}}{\cos x\cos a} \\[6pt] & = \tan a \sec a \end{align}$$