Question

1. evaluate this limit. $lim_{x

ightarrow16}\frac{16 - x}{4-sqrt{x}}cdot\frac{4+sqrt{x}}{4+sqrt{x}}$

Explanation:

Step1: Multiply by conjugate

Multiply the fraction $\frac{16 - x}{4-\sqrt{x}}$ by $\frac{4+\sqrt{x}}{4+\sqrt{x}}$. We get $\lim_{x
ightarrow16}\frac{(16 - x)(4+\sqrt{x})}{(4-\sqrt{x})(4+\sqrt{x})}$.

Step2: Simplify the denominator

Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the denominator $(4-\sqrt{x})(4+\sqrt{x})=16 - x$. So the limit becomes $\lim_{x
ightarrow16}\frac{(16 - x)(4+\sqrt{x})}{16 - x}$.

Step3: Cancel out common factors

Cancel out the common factor $16 - x$ (since $x
eq16$ when taking the limit), we have $\lim_{x
ightarrow16}(4+\sqrt{x})$.

Step4: Substitute the value of x

Substitute $x = 16$ into $4+\sqrt{x}$, we get $4+\sqrt{16}=4 + 4=8$.

Answer:

$8$