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