# How can one find $\lim_{x \to \pm\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$ without graphing?

I'm trying to find both of these:

$$\lim_{x \to +\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$ $$\lim_{x \to -\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$ I know that they end up being $2$ and $-2$ by graphing, but what process can I use to find them without graphing (in the simplest way possible)?

• Have you considered l'Hôpital's rule? – Michael Lee Sep 7 '17 at 22:47

You can apply the good old squeezy theorem.

For $x>1$,

$$\frac1{|x+1|}=\frac1{\sqrt{x^2+2x+1}}<\frac1{\sqrt{x^2+x+1}}<\frac1{\sqrt{x^2}}=\frac1{|x|}$$

For $x<-1$, flip all the inequalities to get

$$\frac1{|x|}<\frac1{\sqrt{x^2+x+1}}<\frac1{|x+1|}$$

• Something is wrong...... for x=2 it's absurd you get $\frac 1 2<\frac1{\sqrt{x^2+x+1}}< \frac 1 3$ – Isham Sep 7 '17 at 23:34
• @Isham Oops, you are right, I have all my inequalities flipped (forgot to flip the inequalities when reciprocating) – Simply Beautiful Art Sep 7 '17 at 23:37

$$\frac{2x+1}{\sqrt{x^2+x+1}}=\frac{x\left(2+\dfrac{1}{x}\right)}{\sqrt{x^2\left(1+\dfrac{1}{x}+\dfrac{1}{x^2}\right)}}=\frac{x\left(2+\dfrac{1}{x}\right)}{|x|\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}=\cdots$$

We look at the highest exponent terms of the polynomials in the numerator and denominator (for the denominator, we look at the polynomial inside the square root), as these are the terms that matter most as $x\to\infty$. $$\frac{2x+1}{\sqrt{x^2+x+1}}\sim\frac{2x}{\sqrt{x^2}}=\frac{2x}{|x|}\longrightarrow\begin{cases} 2\qquad\text{as x\to\infty}\\-2\qquad\text{as x\to -\infty}\end{cases}$$

First recall that $\sqrt{x^2} = x$ if $x\ge0$ and $\sqrt{x^2} = -x$ (which is positive) if $x<0.$

Then treat the two cases, $x>0$ and $x<0,$ separately: \begin{align} \frac{2x+1}{\sqrt{x^2+x+1}} & = \frac{2 + \frac 1 x}{\frac 1 x \sqrt{x^2 + x + 1}} \\[10pt] & = \frac{2+\frac 1 x}{\frac 1 {\sqrt{x^2}}\sqrt{x^2 + x+1}} \quad \text{if } x>0 \\[10pt] & = \frac{2+ \frac 1 x}{\sqrt{1 + \frac 1 x + \frac 1 {x^2}}} \\[10pt] & \to 2 \text{ as } x\to\infty. \\[10pt] \frac{2x+1}{\sqrt{x^2+x+1}} & = \frac{2 + \frac 1 x}{\frac 1 x \sqrt{x^2 + x + 1}} \\[10pt] & = \frac{2+\frac 1 x}{\frac {-1} {\sqrt{x^2}}\sqrt{x^2 + x+1}} \quad \text{if } x<0 \\[10pt] & = \frac{2+ \frac 1 x}{-\sqrt{1 + \frac 1 x + \frac 1 {x^2}}} \\[10pt] & \to -2 \text{ as } x\to\infty. \end{align}

This initially has indeterminate form $\frac{\infty}{\infty}$

Rewrite the expression so that the denominator does not increase without bound (does not go to infinity.)

Factor $\sqrt{x^2} = |x|$ in the denominator. And factor $x$ from the numerator. Simplify and reduce.