​$$\text { If } f(16)=16 \text { and } f^{\prime}(16)=5 \text {, then } \lim _{x \rightarrow 16} \frac{\sqrt{f(x)}-4}{\sqrt{x}-4}=\text { ? }$$​​

Correct option is B

​$$\text{First, substitute } x = 16 \text{ into the expression: } \frac{\sqrt{f(16)} - 4}{\sqrt{16} - 4} = \frac{\sqrt{16} - 4}{4 - 4} = \frac{0}{0}$$

This is an indeterminate form of type 0/0, so we can apply L'Hopital's Rule.

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