Find lim_(x rarr(pi)/(2))(sin(2x))/(cos(x)). Choose 1 answer: (A) (1)/(2) (B) 1 (C) 2 (D) The limit doesn't exist

Question

Find  lim_(x rarr(pi)/(2))(sin(2x))/(cos(x)). Choose 1 answer: (A)  (1)/(2) (B) 1 (C) 2 (D) The limit doesn't exist

Bytelearn · Accepted Answer

Identify Limit: Identify the limit that needs to be evaluated. We need to find the limit of the function sin(2x)/cos(x) as x approaches π/2.Direct Substitution: Direct substitution to check if the limit can be evaluated directly. If we substitute x = π/2 directly into the function, we get sin(π)/cos(π/2). Since sin(π) = 0 and cos(π/2) = 0, we have an indeterminate form of 0/0. This means we cannot find the limit by direct substitution.Apply L'Hôpital's Rule: Apply L'Hôpital's Rule to evaluate the limit. L'Hôpital's Rule states that if the limit of functions f(x) and g(x) as x approaches a value c is an indeterminate form 0/0 or ∞/∞, then the limit of f(x)/g(x) as x approaches c is the same as the limit of their derivatives f'(x)/g'(x), provided this limit exists.Differentiate Functions: Differentiate the numerator and denominator. The derivative of sin(2x) with respect to x is 2cos(2x), and the derivative of cos(x) with respect to x is -sin(x).Apply Derivatives: Apply L'Hôpital's Rule using the derivatives. Now we need to find the limit of (2cos(2x))/(-sin(x)) as x approaches π/2.Substitute Values: Substitute x = π/2 into the derivatives. Substituting x = π/2 into the derivatives, we get (2cos(π))/(-sin(π/2)). Since cos(π) = -1 and sin(π/2) = 1, the expression simplifies to (2 * -1)/(-1) which equals 2.Conclude Limit: Conclude the limit. The limit of sin(2x)/cos(x) as x approaches π/2 is 2.

Find $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin (2 x)}{\cos (x)}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2}$ $\newline$ (B) $1$ $\newline$ (C) $2$ $\newline$ (D) The limit doesn't exist

Full solution

More problems from Find derivatives of logarithmic functions

Related topics

Find lim⁡x→π2sin⁡(2x)cos⁡(x) \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin (2 x)}{\cos (x)} limx→2π​​cos(x)sin(2x)​.\newlineChoose 111 answer:\newline(A) 12 \frac{1}{2} 21​\newline(B) 111\newline(C) 222\newline(D) The limit doesn't exist

Find $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin (2 x)}{\cos (x)}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2}$ $\newline$ (B) $1$ $\newline$ (C) $2$ $\newline$ (D) The limit doesn't exist