Question

Find the following limit

Lim x tengs to 0 1-cos x^2/x^2 sin x^2

Answer 1

$Solution:~ Let~L=\lim_{x\to 0} \frac{1- cos~ x^2 }{x^2 ~sin ~x^2} \\This ~is ~ indeterminant ~form.Therefore~we~apply~L'Hospital ~Rule ~to ~Solve~this ~limit. \\\therefore Apply~L'Hospitals ~Rule \\\therefore~L= \lim_{x\to 0} \frac{2x.sin~ x^2 }{2x.sin~ x^2+2x^3 ~cos ~x^2} \\~~~~~~~~= \lim_{x\to 0} \frac{2x.sin~ x^2 }{2x(sin~ x^2+x^2 ~cos ~x^2)} \\~~~~~~~~= \lim_{x\to 0} \frac{sin~ x^2 }{sin~ x^2+x^2 ~cos ~x^2} \\ which ~is ~again ~ indeterminant ~form. Therefore~we~again~apply ~L'Hospital ~Rule. \\\therefore L=\lim_{x\to 0} \frac{2x ~cos~ x^2 }{4x~cos~ x^2-2x^3 ~sin ~x^2} \\~~~~~~~~=\lim_{x\to 0} \frac{2x ~cos~ x^2 }{2x~(2~cos~ x^2-x^2 ~sin ~x^2)} \\~~~~~~~~=\lim_{x\to 0} \frac{cos~ x^2 }{2~cos~ x^2-x^2 ~sin ~x^2} \\Now ~put ~the ~limit ~x=0 \\\therefore L=\lim_{x\to 0} \frac{cos~ 0^2 }{2~cos~ 0^2-0^2 ~sin ~0^2}=\frac{1}{2} \\\lim_{x\to 0} \frac{1- cos~ x^2 }{x^2 ~sin ~x^2}=\frac{1}{2}$