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3 years ago

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Please help me ......

1 answer:

Elan Coil [88]3 years ago

4 0

Egudbschsien eiodnebevieneb

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

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$\displaystyle\lim_{x\to3^-}\frac x{\sqrt{x^2-9}}=\lim_{x\to3^-}\frac x{\sqrt{x^2}\sqrt{1-\frac9{x^2}}}=\lim_{x\to3^-}\frac x{|x|\sqrt{1-\frac9{x^2}}}$

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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