Answer to Question #190416 in Calculus for Nikhil Rawat

Question

lim (x.cosx- sinx / x^2.sinx)

x→0

Answer 1

$Lim_{x→0}​ \frac{xcosx−sinx}{x^2 sinx}\hspace{1.3cm}(\frac{0}{0} form)\newline​ \text{By L'Hospital's rule,}\newline =Lim_{x→0}​ \frac{xcosx−sinx}{x^2 (\frac{sinx}{x})x}\newline =Lim_{x→0}​ \frac{xcosx−sinx}{x^3}\hspace{1cm}(lim\frac{sinx}{x}=1)\newline \text{Differentiate numerator and denominator}\newline =Lim_{x→0}​ \frac{-xsinx+cosx−cosx}{3x^2}\newline =Lim_{x→0}​ \frac{-xsinx}{3x^2}\newline =Lim_{x→0}​ \frac{-x}{3x}(\frac{sinx}{x})\newline =Lim_{x→0}​ \frac{-1}{3}(\frac{sinx}{x})\hspace{1cm}(lim\frac{sinx}{x}=1)\newline =\frac{-1}{3}$