The value of lim_{x to 0} left( frac{x^2 sin^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We use the Taylor series expansion for $\sin x$ near $x = 0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$.Then, $\sin^2 x = (x - \frac{x^3}{6} + O(x^5))

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) We use the Taylor series expansion for $\sin x$ near $x = 0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$.Then, $\sin^2 x = (x - \frac{x^3}{6} + O(x^5))^2 = x^2 - 2(x)(\frac{x^3}{6}) + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$.Substituting this into the denominator: $x^2 - \sin^2 x = x^2 - (x^2 - \frac{x^4}{3} + O(x^6)) = \frac{x^4}{3} + O(x^6)$.Now, substitute this into the limit expression: $\lim_{x 	o 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} = \lim_{x 	o 0} \frac{x^2 (x^2 + O(x^4))}{\frac{x^4}{3} + O(x^6)}$.$= \lim_{x 	o 0} \frac{x^4 + O(x^6)}{\frac{x^4}{3} + O(x^6)} = \frac{1}{1/3} = 3$.

The value of $\lim_{x \to 0} \left( \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} \right)$ is:

Similar Questions

If ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + \dots - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ then $\mathop {\lim }\limits_{n \to \infty } {x_n}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

For each $t \in \mathbb{R}$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$ is:

$\lim _{x \rightarrow 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin ^2 x} = $

$\lim\limits _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module