BapnaHT 26. B) lim _(xarrow -8)(3x+24)/(sin(2x+16))

To find the limit of the given expression as \( x \) approaches \(-8\), we can use the fact that \(\sin(2x+16)\) approaches 0 as \(x\) approaches \(-8\). <br /><br />We can rewrite the expression as:<br /><br />\[<br />\lim_{x \to -8} \frac{3x+24}{\sin(2x+16)}<br />\]<br /><br />As \(x\) approaches \(-8\), the numerator \(3x+24\) approaches 0, and the denominator \(\sin(2x+16)\) approaches 0. This is an indeterminate form of type \(\frac{0}{0}\), so we can apply L'Hôpital's Rule.<br /><br />Taking the derivative of the numerator and denominator with respect to \(x\), we get:<br /><br />\[<br />\lim_{x \to -8} \frac{3}{2\cos(2x+16)}<br />\]<br /><br />Now, as \(x\) approaches \(-8\), the denominator \(\cos(2x+16)\) approaches \(\cos(0)\), which is 1. Therefore, the limit is:<br /><br />\[<br />\lim_{x \to -8} \frac{3}{2\cos(2x+16)} = \frac{3}{2 \cdot 1} = \frac{3}{2}<br />\]<br /><br />So, the correct answer is \(\frac{3}{2}\).