For what value of k is the function f(x)={[(sin x+x cos x)/(x)","x!=0],[3k","x=0]:} is continuous at x=0 ?

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For what value of  k is the function  f(x)={[(sin x+x cos x)/(x)","x!=0],[3k","x=0]:} is continuous at  x=0 ?

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Define Limit Approach: To determine the value of k for which the function f(x) is continuous at x=0, we need to ensure that the limit of f(x) as x approaches 0 from the left and right equals the value of f(x) at x=0.Calculate Right Limit: First, we find the limit of the function as x approaches 0 from the right (x > 0). We use the given piece of the function for x ≠ 0, which is (sin x + x cos x) / x. lim (x -> 0+) [(sin x + x cos x) / x] = lim (x -> 0+) [sin x / x + cos x]Find Function Value: We know that lim (x -> 0) [sin x / x] = 1 and lim (x -> 0) [cos x] = cos(0) = 1. So, lim (x -> 0+) [(sin x + x cos x) / x] = 1 + 1 = 2.Ensure Continuity: Now, we need to find the value of the function at x=0, which is given as f(0) = 3k.Set Limit Equal: For the function to be continuous at x=0, the limit as x approaches 0 from the right must equal the value of the function at x=0. So, we set the limit equal to the value of f(0): 2 = 3k.Solve for k: Solving for k, we get k = 2 / 3.

For what value of $k$ is the function $f(x)=\begin{cases} \frac{\sin x+x\cos x}{x}, & x\neq 0, \ 3k, & x=0 \end{cases}$ is continuous at $x=0$ ?

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For what value of $k$ is the function $f(x)=\begin{cases} \frac{\sin x+x\cos x}{x}, & x\neq 0, \ 3k, & x=0 \end{cases}$ is continuous at $x=0$ ?