Find $\lim_{x\to a}\frac{\tan x-\tan a}{\ln x-\ln a}$

Question

$$\lim_{x\to a}\frac{\tan x-\tan a}{\ln x-\ln a}$$ Solve that limit while $a$ is unknown.

I have tried applying l’Hopital rule however it didn’t seem to work. I have tried simplifying it but I’m still very confused on how are you able to take a limit to an unknown variable.

Any explanation is welcome. Thanks!

Answer 1

Let $h$ be small. Then $$\tan(a+h) - \tan(a)=\sec^2(a)\cdot h + o(h)$$ and $$\ln(a+h) - \ln(a) = \ln \left(1+\frac{h}{a}\right) = \frac{h}{a}+o(h)$$ hence $$f(a+h) = \frac{\sec^2(a)\cdot h + o(h)}{h/a + o(h)}=a\sec^2(a) +o(1)$$ so the desired limit is $$a\sec^2(a)$$

Answer 2

Hint:

$$\dfrac{\tan x-\tan a}{\ln x-\ln a}=\dfrac{\sin(x-a)}{\cos x\cos a\ln\left(1+\dfrac{x-a}a\right)}$$

$$=\dfrac1{\cos x\cos a}\cdot\dfrac{\sin(x-a)}{x-a}\cdot\dfrac1{ \dfrac{a\cdot\ln\left(1+\dfrac{x-a}a\right)}{\dfrac{x-a}a}}$$

Now use $\lim_{h\to0}\dfrac{\sin h}h=1$ and $\lim_{k\to0}\dfrac{\ln(1+k)}k=1$

Answer 3

Hint

$$\frac{\tan(x)-\tan(a)}{\ln(x)-\ln(a)}=\frac{\tan(x)-\tan(a)}{x-a}\frac{1}{\frac{\ln(x)-\ln(a)}{x-a}}.$$

Answer 4

Hint: Use L'Hopital rule: $$\lim_{x\to a}\frac{\tan x-\tan a}{\ln x-\ln a}=\lim_{x\to a}\frac{1+\tan^2 x}{\frac1x}$$