# Find the limit: lim_{xrightarrowinfty}frac{sqrt{9x+1}}{sqrt{x+1}}

Question
Limits and continuity
Find the limit:
$$\lim_{x\rightarrow\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}$$

2021-01-08
To find the liimit,
$$\lim_{x\rightarrow\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}$$
$$\lim_{x\rightarrow\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}=\lim_{x\rightarrow\infty}\sqrt\frac{9x+1}{x+1}$$
$$=\lim_{x\rightarrow\infty}\sqrt\frac{x(9+\frac{1}{x})}{x(1+\frac{1}{x})}$$
$$=\lim_{x\rightarrow\infty}\sqrt\frac{(9+\frac{1}{x})}{(1+\frac{1}{x})}$$
$$=\lim_{x\rightarrow\infty}\sqrt\frac{(9+0)}{(1+0)}\ \ \ as\ x\rightarrow\infty,\frac{1}{x}\rightarrow0$$
$$=\sqrt9$$
$$=3$$
Therefore, $$\lim_{x\rightarrow\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}=3$$

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