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

Question
Limits and continuity
asked 2020-10-18
Find the limits
\(\lim_{x\rightarrow9}\frac{\sin(\sqrt x-3)}{x-9}\)

Answers (1)

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 \(\frac{1}{6}\)
0

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