evaluate the limit: $lim_{x\rightarrow0}\frac{sqrt{2x + 9}-3}{x}$
evaluate the limit: $lim_{x\rightarrow0}\frac{sqrt{2x + 9}-3}{x}$
Answer
Explanation:
Step1: Rationalize the numerator
Multiply the fraction by $\frac{\sqrt{2x + 9}+3}{\sqrt{2x + 9}+3}$. [ \begin{align*} &\lim_{x\rightarrow0}\frac{\sqrt{2x + 9}-3}{x}\times\frac{\sqrt{2x + 9}+3}{\sqrt{2x + 9}+3}\ =&\lim_{x\rightarrow0}\frac{(\sqrt{2x + 9})^2-3^2}{x(\sqrt{2x + 9}+3)}\ =&\lim_{x\rightarrow0}\frac{2x + 9 - 9}{x(\sqrt{2x + 9}+3)}\ =&\lim_{x\rightarrow0}\frac{2x}{x(\sqrt{2x + 9}+3)} \end{align*} ]
Step2: Simplify the fraction
Cancel out the common - factor $x$ in the numerator and denominator. [ \begin{align*} &\lim_{x\rightarrow0}\frac{2x}{x(\sqrt{2x + 9}+3)}\ =&\lim_{x\rightarrow0}\frac{2}{\sqrt{2x + 9}+3} \end{align*} ]
Step3: Evaluate the limit
Substitute $x = 0$ into the simplified function. [ \begin{align*} &\frac{2}{\sqrt{2\times0+9}+3}\ =&\frac{2}{\sqrt{9}+3}\ =&\frac{2}{3 + 3}\ =&\frac{1}{3} \end{align*} ]
Answer:
$\frac{1}{3}$