Limit question

**Monster32432421** · May 1st 2010, 08:02 PM

Determine Lim x->Infinity Sqrt[4 x^2 + 7 x - 2] - (2 x + 1)
I know the answer, but i dont know the steps to get to it..

**simplependulum** · May 1st 2010, 08:17 PM

Originally Posted by Monster32432421

Determine Lim x->Infinity Sqrt[4 x^2 + 7 x - 2] - (2 x + 1)
I know the answer, but i dont know the steps to get to it..

Try to rationalise the numerator .

**tonio** · May 1st 2010, 08:21 PM

Originally Posted by Monster32432421

Determine Lim x->Infinity Sqrt[4 x^2 + 7 x - 2] - (2 x + 1)
I know the answer, but i dont know the steps to get to it..

Write $\left[\sqrt{4x^2+7x-2}-(2x+1)\right]\,\frac{\sqrt{4x^2+7x-2}+(2x+1)}{\sqrt{4x^2+7x-2}+(2x+1)}$ $=\frac{3x-3}{\sqrt{4x^2+7x-2}+(2x+1)}$ , and now multiply the right hand by $\frac{1/x}{1/x}$ and do

a little algebra + some arithmetic of limits.

The limit is $\frac{3}{4}=0.75$

Tonio

**simplependulum** · May 1st 2010, 08:23 PM

also a little care , you may make mistakes very easily .

**Monster32432421** · May 1st 2010, 08:28 PM

Originally Posted by tonio

Write $\left[\sqrt{4x^2+7x-2}-(2x+1)\right]\,\frac{\sqrt{4x^2+7x-2}+(2x+1)}{\sqrt{4x^2+7x-2}+(2x+1)}$ $=\frac{3x-3}{\sqrt{4x^2+7x-2}+(2x+1)}$ , and now multiply the right hand by $\frac{1/x}{1/x}$ and do

a little algebra + some arithmetic of limits.

The limit is $\frac{3}{4}=0.75$

Tonio

thanks

**maddas** · May 1st 2010, 08:30 PM

${\sqrt{4x^2+7x-2}\over x} = \sqrt{4x^2+7x-2\over x^2} = \sqrt{4 + \frac7x -\frac2{x^2}}$

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