evaluate lim h→0 (sec(x + h) - sec x)/h - csc²x - sec x tan x sec x tan x csc²x

Explanation:

Step1: Recall secant - definition

$\sec(x)=\frac{1}{\cos(x)}$, so $\lim_{h\rightarrow0}\frac{\sec(x + h)-\sec(x)}{h}=\lim_{h\rightarrow0}\frac{\frac{1}{\cos(x + h)}-\frac{1}{\cos(x)}}{h}$

Step2: Combine fractions in numerator

$\lim_{h\rightarrow0}\frac{\frac{\cos(x)-\cos(x + h)}{\cos(x)\cos(x + h)}}{h}=\lim_{h\rightarrow0}\frac{\cos(x)-\cos(x + h)}{h\cos(x)\cos(x + h)}$

Step3: Use the cosine addition formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$

$\cos(x + h)=\cos x\cos h-\sin x\sin h$, then $\cos(x)-\cos(x + h)=\cos x-( \cos x\cos h-\sin x\sin h)=\cos x(1 - \cos h)+\sin x\sin h$ So the limit becomes $\lim_{h\rightarrow0}\frac{\cos x(1 - \cos h)+\sin x\sin h}{h\cos(x)\cos(x + h)}$

Step4: Split the fraction

$\lim_{h\rightarrow0}\left(\frac{\cos x(1 - \cos h)}{h\cos(x)\cos(x + h)}+\frac{\sin x\sin h}{h\cos(x)\cos(x + h)}\right)$

Step5: Recall limit formulas $\lim_{h\rightarrow0}\frac{1 - \cos h}{h}=0$ and $\lim_{h\rightarrow0}\frac{\sin h}{h}=1$

For the first - term $\lim_{h\rightarrow0}\frac{\cos x(1 - \cos h)}{h\cos(x)\cos(x + h)}=\cos x\cdot\lim_{h\rightarrow0}\frac{1 - \cos h}{h}\cdot\lim_{h\rightarrow0}\frac{1}{\cos(x)\cos(x + h)} = 0$ For the second - term $\lim_{h\rightarrow0}\frac{\sin x\sin h}{h\cos(x)\cos(x + h)}=\sin x\cdot\lim_{h\rightarrow0}\frac{\sin h}{h}\cdot\lim_{h\rightarrow0}\frac{1}{\cos(x)\cos(x + h)}$ Since $\lim_{h\rightarrow0}\frac{\sin h}{h}=1$ and $\lim_{h\rightarrow0}\cos(x + h)=\cos x$, the second - term is $\frac{\sin x}{\cos^{2}x}=\sec x\tan x$

Answer:

$\sec x\tan x$