If f(x)= ) x^2sin((1)/(x)),&ifxneq 0 6-2k,&ifx=0 is continuous at x=0 then the value of k=

To determine the value of \( k \) that makes the function \( f(x) \) continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).<br /><br />First, let's find the limit of \( f(x) \) as \( x \) approaches 0:<br /><br />\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \]<br /><br />Since \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1, we can use the Squeeze Theorem to evaluate this limit. We know that:<br /><br />\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]<br /><br />Multiplying through by \( x^2 \), we get:<br /><br />\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]<br /><br />Taking the limit of each part as \( x \) approaches 0, we get:<br /><br />\[ \lim_{x \to 0} -x^2 = 0 \]<br />\[ \lim_{x \to 0} x^2 = 0 \]<br /><br />Therefore, by the Squeeze Theorem:<br /><br />\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]<br /><br />For \( f(x) \) to be continuous at \( x = 0 \), we must have:<br /><br />\[ f(0) = \lim_{x \to 0} f(x) \]<br /><br />Given that \( f(0) = 6 - 2k \), we set this equal to the limit we found:<br /><br />\[ 6 - 2k = 0 \]<br /><br />Solving for \( k \):<br /><br />\[ 6 = 2k \]<br />\[ k = 3 \]<br /><br />Thus, the value of \( k \) that makes the function \( f(x) \) continuous at \( x = 0 \) is \( \boxed{3} \).