Given

\(\lim_{x\rightarrow0}f(x)=\frac{1}{2}\) and \(\lim g(x)=\sqrt2\)

we have to find

\(f(x)\frac{\cos x}{x-1}\)

we know

\(\lim_{x\rightarrow0}f(x)\cdot g(x)=[\lim_{x\rightarrow0}f(x)][\lim_{x\rightarrow0}g(x)]\)

and \(\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0} \frac{f(x)}{\lim_{x\rightarrow0}g(x)}\)

\(\lim_{x\rightarrow0}\frac{f(x)\cos x}{x-1}=[\lim_{x\rightarrow0} f(x)]\cdot[\lim_{x\rightarrow0}\frac{\cos x}{x-1}]\)

\(=\frac{\frac{1}{2}\lim_{x\rightarrow0}\cos x}{\lim_{x\rightarrow0}(x-1)}\)

\(=\frac{1}{2}\cdot\frac{1}{-1}\)

\(=-\frac{1}{2}\)

\(\lim_{x\rightarrow0}f(x)=\frac{1}{2}\) and \(\lim g(x)=\sqrt2\)

we have to find

\(f(x)\frac{\cos x}{x-1}\)

we know

\(\lim_{x\rightarrow0}f(x)\cdot g(x)=[\lim_{x\rightarrow0}f(x)][\lim_{x\rightarrow0}g(x)]\)

and \(\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0} \frac{f(x)}{\lim_{x\rightarrow0}g(x)}\)

\(\lim_{x\rightarrow0}\frac{f(x)\cos x}{x-1}=[\lim_{x\rightarrow0} f(x)]\cdot[\lim_{x\rightarrow0}\frac{\cos x}{x-1}]\)

\(=\frac{\frac{1}{2}\lim_{x\rightarrow0}\cos x}{\lim_{x\rightarrow0}(x-1)}\)

\(=\frac{1}{2}\cdot\frac{1}{-1}\)

\(=-\frac{1}{2}\)