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Alexus [3.1K]
3 years ago
5

As the limit of x goes to 3 from the left find the limit of (x/ sqrt x^2 -9)

Mathematics
1 answer:
Dimas [21]3 years ago
8 0
\displaystyle\lim_{x\to3^-}\frac x{\sqrt{x^2-9}}=\lim_{x\to3^-}\frac x{\sqrt{x^2}\sqrt{1-\frac9{x^2}}}=\lim_{x\to3^-}\frac x{|x|\sqrt{1-\frac9{x^2}}}

Since x>0, you have |x|=x and the limand reduces to

\displaystyle\lim_{x\to3^-}\frac1{\sqrt{1-\frac9{x^2}}}

For x>3, you have \sqrt{1-\frac9{x^2}}>0 since \dfrac9{x^2} will always be smaller than 1, which means \dfrac1{\sqrt{1-\frac9{x^2}}}\to+\infty
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