Solve: $\lim\limits_{y\mapsto 0} \frac{(x+y)\sec(x+y)-x\sec(x)}{y}$

Question

How can I solve the following limit:

$$\lim\limits_{y\mapsto 0} \frac{(x+y)\sec(x+y)-x\sec(x)}{y}$$

What I tried:

$$=\lim\limits_{y\mapsto 0} \frac{(x+y)\sec(x+y)}{y} - \lim\limits_{y\mapsto 0}\frac{x\sec(x)}{y}$$

$$\frac{x\sec(x)}{0}-\frac{x\sec(x)}{0}$$

$$=\infty-\infty$$

Which is an indeterminate form.

I have seen someone asking the same question which has been closed. However I didn't get any solution that that I could comprehend well.

Answer 1

$$\lim\limits_{y\mapsto 0} \frac{(x+y)\sec(x+y)-x\sec x}{y}=\dfrac{d(x\sec x)}{dx}=?$$

Alternatively,

$$\lim_{h\to0}\dfrac{(x+h)\sec(x+h)-x\sec x}h=\lim_{h\to0}\sec(x+h)+x\lim_{h\to0}\dfrac{\sec(x+h)-\sec x}h=?$$

$$\dfrac{\sec(x+h)-\sec x}h=\dfrac{\cos x-\cos(x+h)}{h\cos x\cos(x+h)}=\cos x\cos(x+h)\cdot\dfrac{\sin\dfrac h2\sin\left(x+\dfrac h2\right)}{\dfrac h2}=?$$

Answer 2

First regarding your attempt, we can't split limit like that. We can only split the limit of $\lim_{y \to y_0 } f(y) + g(y) = \lim_{y \to y_0 } f(y) + \lim_{y \to y_0 }g(y)$ if the limits on the right exists.

Upon recognizing it as a derivative, you can use the product rules.

$$\lim\limits_{y\mapsto 0} \frac{(x+y)\sec(x+y)-x\sec x}{y}= \frac{d(x \sec x)}{dx} = x\frac{d\sec x}{dx} + \sec x\frac{dx}{dx}$$

Answer 3

this limit is the definition of derivativ $x \sec x$ therefor $$(x \sec x)'= \sec x-x \tan x \sec x $$