Question

Evaluate the Given limit: $\lim_{x\rightarrow 0}$ $\frac{cos^2x-1}{cos\,x-1}$

Answer 1

$\lim_{x\rightarrow 0}$ $\frac{cos^2x-1}{cos\,x-1}$
At x = 0, the value of the given function takes the form 0/0.
Now,
=$\lim_{x\rightarrow 0}$ $\frac{cos^2x-1}{cos\,x-1}$ =$\lim_{x\rightarrow 0}$ $\frac{1-2sin^2x-1}{1-2sin^2\frac{x}{2}-1}$ [cosx = 1 - 2sin2$\frac{x}{2}$]
=$\lim_{x\rightarrow 0}$ $\frac{sin^2x}{\frac{sin^2x}{2}}$ = $\lim_{x\rightarrow 0}$ ($\frac{sin^2x}{2}$)\times$$\frac{x^2}{\frac{sin^2x}{2}}$$\times$$\frac{2}{4}
=$\frac{4\lim_{x\rightarrow 0}(\frac{sin^2x}{x^2})}{\lim_{x\rightarrow 0}(\frac{sin^2\frac{x}{2}}{\frac{x}{2}^2})}$
=4 \times$$\frac{12}{12} [$\lim_{y\rightarrow 0}$ $\frac{sin\,y}{y}$ = 1]
=4