Question

Evaluate limx→[infinity](2x−√(4x2+5x+5)).

Answer 1

-5/4

Explanation:

First, rewrite the function:

`\displaystyle 2x-√(4x^2+5x+5)\\\\=((2x-√(4x^2+5x+5))(2x+√(4x^2+5x+5)))/(2x+√(4x^2+5x+5))\\\\=(4x^2-(4x^2+5x+5))/(2x+√(4x^2+5x+5))\\\\=(4x^2-4x^2-5x-5)/(2x+√(4x^2+5x+5))\\\\=(-5x-5)/(2x+√(4x^2+5x+5))\\\\=\frac{-5x-5}{2x+√(x^2)\sqrt{4+(5)/(x)+(5)/(x^2)}}`

Since we are taking the limit of the function to positive infinity, then we say that
`√(x^2)=x`:

`\displaystyle \frac{-5x-5}{2x+x\sqrt{4+(5)/(x)+(5)/(x^2)}}\\\\=\frac{-5-(5)/(x)}{2+\sqrt{4+(5)/(x)+(5)/(x^2)}}`

Now the limit becomes easier to evaluate:

`\displaystyle \lim_(x \to \infty) \frac{-5-(5)/(x)}{2+\sqrt{4+(5)/(x)+(5)/(x^2)}}\\\\=\frac{-5-(5)/(\infty)}{2+\sqrt{4+(5)/(\infty)+(5)/(\infty^2)}}\\\\=(-5-0)/(2+√(4+0+0))\\\\=(-5)/(2+√(4))\\\\=(-5)/(2+2)\\\\=-(5)/(4)`