question evaluate the limit: $lim_{x \to 9}\frac{x - 9}{sqrt{x}-3}$
question evaluate the limit: $lim_{x \to 9}\frac{x - 9}{sqrt{x}-3}$
Answer
Answer:
6
Explanation:
Step1: Rationalize the denominator
Multiply numerator and denominator by $\sqrt{x}+3$: [ \begin{align*} \lim_{x\rightarrow9}\frac{x - 9}{\sqrt{x}-3}&=\lim_{x\rightarrow9}\frac{(x - 9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\ \end{align*} ]
Step2: Simplify the denominator
Use the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$. Here $a=\sqrt{x}$ and $b = 3$, so $(\sqrt{x}-3)(\sqrt{x}+3)=x - 9$. [ \begin{align*} \lim_{x\rightarrow9}\frac{(x - 9)(\sqrt{x}+3)}{x - 9}&=\lim_{x\rightarrow9}(\sqrt{x}+3) \end{align*} ]
Step3: Evaluate the limit
Substitute $x = 9$ into $\sqrt{x}+3$: [ \begin{align*} \sqrt{9}+3&=3 + 3=6 \end{align*} ]