record: 196 score: 196 evaluate lim h→0 (tan(x + h) - tan x)/h csc²x sec²x - csc²x - sec²x

Answer:

B. $\sec^{2}x$

Explanation:

Step1: Recall the tangent - addition formula

$\tan(A + B)=\frac{\tan A+\tan B}{1 - \tan A\tan B}$, so $\tan(x + h)=\frac{\tan x+\tan h}{1 - \tan x\tan h}$.

Step2: Substitute into the limit

$\lim_{h\rightarrow0}\frac{\tan(x + h)-\tan x}{h}=\lim_{h\rightarrow0}\frac{\frac{\tan x+\tan h}{1 - \tan x\tan h}-\tan x}{h}$.

Step3: Simplify the numerator

$\frac{\tan x+\tan h}{1 - \tan x\tan h}-\tan x=\frac{\tan x+\tan h-\tan x(1 - \tan x\tan h)}{1 - \tan x\tan h}=\frac{\tan x+\tan h-\tan x+\tan^{2}x\tan h}{1 - \tan x\tan h}=\frac{\tan h+\tan^{2}x\tan h}{1 - \tan x\tan h}=\frac{\tan h(1 + \tan^{2}x)}{1 - \tan x\tan h}$.

Step4: Rewrite the limit

$\lim_{h\rightarrow0}\frac{\tan(x + h)-\tan x}{h}=\lim_{h\rightarrow0}\frac{\frac{\tan h(1 + \tan^{2}x)}{1 - \tan x\tan h}}{h}=\lim_{h\rightarrow0}\frac{\tan h}{h}\cdot\frac{1+\tan^{2}x}{1 - \tan x\tan h}$.

Step5: Use the well - known limit $\lim_{h\rightarrow0}\frac{\tan h}{h}=1$

As $h\rightarrow0$, $\lim_{h\rightarrow0}\frac{\tan h}{h}=1$ and $\lim_{h\rightarrow0}(1 - \tan x\tan h)=1$. Also, $1+\tan^{2}x=\sec^{2}x$. So the limit is $\sec^{2}x$.