# Find the limits lim_{xrightarrow9}frac{sin(sqrt x-3)}{x-9}

Question
Limits and continuity
Find the limits
$$\lim_{x\rightarrow9}\frac{\sin(\sqrt x-3)}{x-9}$$

2020-10-19
Find the limit,
$$\lim_{x\rightarrow9}\frac{\sin(\sqrt x-3)}{x-9}$$
Here we have indeterminate form $$\frac{0}{0}$$
So, apply the L'Hospital rule,
Thus differentiate the numerator and denominator then take the provided limit,
$$\lim_{x\rightarrow9}\frac{\sin(\sqrt x-3)}{x-9}=\lim_{x\rightarrow9}\frac{\cos(\sqrt x-3)\cdot\frac{1}{2\sqrt x}}{1-0}$$
$$=\lim_{x\rightarrow9}\frac{\frac{\cos(\sqrt x-3)}{2\sqrt x}}{1}$$
$$=\lim_{x\rightarrow9}\frac{\cos(\sqrt x-3)}{2\sqrt x}$$
Now take the limit $$x\rightarrow9,$$
$$\lim_{x\rightarrow9}\frac{\sin(\sqrt x-3)}{x-9}=\frac{\cos(\sqrt 9-3)}{2\sqrt9}$$
$$=\frac{\cos(3-3)}{2\cdot3}$$
$$=\frac{\cos0}{6}$$
$$=\frac{1}{6}$$
Hence the limit is $$\frac{1}{6}$$

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