Question

1. evaluate

lim_{x
ightarrow16}\frac{16 - x}{4-sqrt{x}}

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $4 + \sqrt{x}$.
\[

$$\begin{align*} \lim_{x ightarrow16}\frac{16 - x}{4-\sqrt{x}}&=\lim_{x ightarrow16}\frac{(16 - x)(4+\sqrt{x})}{(4-\sqrt{x})(4 + \sqrt{x})}\\ \end{align*}$$

\]

Step2: Simplify the denominator using difference - of - squares

Since $(a - b)(a + b)=a^{2}-b^{2}$, then $(4-\sqrt{x})(4+\sqrt{x})=16 - x$.
\[

$$\begin{align*} \lim_{x ightarrow16}\frac{(16 - x)(4+\sqrt{x})}{(4-\sqrt{x})(4 + \sqrt{x})}&=\lim_{x ightarrow16}\frac{(16 - x)(4+\sqrt{x})}{16 - x}\\ \end{align*}$$

\]

Step3: Cancel out the common factor

Cancel out the common factor $16 - x$ (since $x
eq16$ when taking the limit).
\[

$$\begin{align*} \lim_{x ightarrow16}\frac{(16 - x)(4+\sqrt{x})}{16 - x}&=\lim_{x ightarrow16}(4+\sqrt{x}) \end{align*}$$

\]

Step4: Evaluate the limit

Substitute $x = 16$ into $4+\sqrt{x}$.
\[

$$\begin{align*} \lim_{x ightarrow16}(4+\sqrt{x})&=4+\sqrt{16}\\ &=4 + 4\\ &=8 \end{align*}$$

\]

Answer:

$8$