Evaluate lim x-> 0 (sin x)/x

Limit Property Explanation: We will use the limit property that if the limit of the numerator and the limit of the denominator both exist and the limit of the denominator is not zero, then the limit of the fraction is the quotient of the limits. However, in this case, both the numerator and the denominator approach 0, which is an indeterminate form. We need to use L'Hôpital's Rule, which states that if the limit of f(x)/g(x) as x approaches a value c is an indeterminate form 0/0 or ∞/∞, then the limit is the same as the limit of f'(x)/g'(x) as x approaches c, provided that the derivatives exist and the limit of f'(x)/g'(x) exists or is ∞ or -∞.Apply L'Hôpital's Rule: We will apply L'Hôpital's Rule to the limit of (sin x)/x as x approaches 0. We need to find the derivatives of the numerator and the denominator with respect to x. The derivative of sin x with respect to x is cos x. The derivative of x with respect to x is 1.Find Derivatives: Now we will take the limit of the derivatives as x approaches 0. lim (x -> 0) (cos x) / 1 Since the derivative of the denominator is a constant (1), we only need to evaluate the derivative of the numerator at x = 0.Evaluate Derivatives: Evaluating the derivative of the numerator at x = 0 gives us cos(0), which is equal to 1. Therefore, the limit of (cos x) / 1 as x approaches 0 is 1/1, which simplifies to 1.Calculate Final Limit: We have found that the limit of (sin x)/x as x approaches 0 is equal to the limit of (cos x)/1 as x approaches 0, which we have determined to be 1.

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Evaluate $\lim_{x\to 0} \frac{\sin x}{x}$

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Q. Evaluate

\lim_{x\to 0} \frac{\sin x}{x}

Limit Property Explanation: We will use the limit property that if the limit of the numerator and the limit of the denominator both exist and the limit of the denominator is not zero, then the limit of the fraction is the quotient of the limits. However, in this case, both the numerator and the denominator approach $0$ , which is an indeterminate form. We need to use L'Hôpital's Rule, which states that if the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches a value $c$ is an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then the limit is the same as the limit of $\frac{f'(x)}{g'(x)}$ as $x$ approaches $c$ , provided that the derivatives exist and the limit of $\frac{f'(x)}{g'(x)}$ exists or is $\frac{f(x)}{g(x)}$ $0$ or $\frac{f(x)}{g(x)}$ $1$ .
Apply L'Hôpital's Rule: We will apply L'Hôpital's Rule to the limit of $\frac{\sin x}{x}$ as $x$ approaches $0$ . We need to find the derivatives of the numerator and the denominator with respect to $x$ . $\newline$ The derivative of $\sin x$ with respect to $x$ is $\cos x$ . $\newline$ The derivative of $x$ with respect to $x$ is $1$ .
Find Derivatives: Now we will take the limit of the derivatives as $x$ approaches $0$ . $\lim_{x \to 0} \frac{\cos x}{1}$ Since the derivative of the denominator is a constant ( $1$ ), we only need to evaluate the derivative of the numerator at $x = 0$ .
Evaluate Derivatives: Evaluating the derivative of the numerator at $x = 0$ gives us $\cos(0)$ , which is equal to $1$ . Therefore, the limit of $(\cos x) / 1$ as $x$ approaches $0$ is $1/1$ , which simplifies to $1$ .
Calculate Final Limit: We have found that the limit of $(\sin x)/x$ as $x$ approaches $0$ is equal to the limit of $(\cos x)/1$ as $x$ approaches $0$ , which we have determined to be $1$ .