Thread: Not L'Hopital rule
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Old October 5th, 2007, 12:29 PM
Soroban Soroban is offline
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Hello, Joyce!

Another approach . . .

Quote:
\lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+2}}

We have: .\lim_{x\to\infty}\sqrt{\frac{9x+1}{x+2}}

Divide top and bottom by x\!:\;\;\lim_{x\to\infty}\sqrt{\frac{9 + \frac{1}{x}}{1 + \frac{2}{x}}} \;=\;\sqrt{\frac{9+0}{1+0}} \:=\:\sqrt{9} \;=\;3
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