Thread: limits
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Old December 1st, 2008, 08:25 PM
Rapha Rapha is offline
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Hi Faz

Quote:
Originally Posted by Faz View Post
find the following limits


lim ->infinite ((x^2/x+5)-x)
Do you mean lim_{x \to \infty } \frac{x^2}{x+5}-x?

Then

lim_{x \to \infty } \frac{x^2}{x+5}-\frac{x(x+5)}{x+5}

= lim_{x \to \infty } \frac{x^2- x(x+5)}{x+5}

= lim_{x \to \infty } \frac{x^2- x^2 - 5x)}{x+5}


= lim_{x \to \infty } \frac{-5x}{x+5} = lim_{x \to \infty } \frac{-5}{1+\frac{5}{x}} = - 5
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