The value of operatorname{Lt}_{x rightarrow 0 | vedclass

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(B) We need to evaluate the limit: $\operatorname{Lt}_{x \rightarrow 0} \frac{\sin^2 x + \cos x - 1}{x^2}$.Using the identity $\sin^2 x = 1 - \cos^2 x

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) We need to evaluate the limit: $\operatorname{Lt}_{x ightarrow 0} \frac{\sin^2 x + \cos x - 1}{x^2}$.Using the identity $\sin^2 x = 1 - \cos^2 x$, the expression becomes:$\operatorname{Lt}_{x ightarrow 0} \frac{1 - \cos^2 x + \cos x - 1}{x^2} = \operatorname{Lt}_{x ightarrow 0} \frac{\cos x - \cos^2 x}{x^2}$.Factor out $\cos x$:$\operatorname{Lt}_{x ightarrow 0} \frac{\cos x(1 - \cos x)}{x^2} = \operatorname{Lt}_{x ightarrow 0} \cos x \cdot \operatorname{Lt}_{x ightarrow 0} \frac{1 - \cos x}{x^2}$.We know that $\operatorname{Lt}_{x ightarrow 0} \cos x = 1$ and $\operatorname{Lt}_{x ightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.Therefore, the limit is $1 	imes \frac{1}{2} = \frac{1}{2}$.

The value of $\operatorname{Lt}_{x \rightarrow 0} \frac{\sin^2 x + \cos x - 1}{x^2}$ is

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