# How do you find the limit of (sqrt(x^2+1))/(3x-1) as x approaches infinity?

Feb 22, 2015

for $x \rightarrow + \infty$, the limit is $\frac{1}{3}$,

For $x \rightarrow - \infty$, the limit is $- \frac{1}{3}$.

This is because:

${\lim}_{x \rightarrow \pm \infty} \frac{\sqrt{{x}^{2} + 1}}{3 x - 1} = {\lim}_{x \rightarrow \pm \infty} \frac{\sqrt{{x}^{2}}}{3 x} =$

$= {\lim}_{x \rightarrow \pm \infty} | x \frac{|}{3 x} = \left(1\right)$

For $x \rightarrow + \infty$, $| x | = x$,

for $x \rightarrow - \infty$, $| x | = - x$.

So:

$\left(1\right) = {\lim}_{x \rightarrow + \infty} \frac{x}{3 x} = \frac{1}{3}$,

$\left(1\right) = {\lim}_{x \rightarrow - \infty} \frac{- x}{3 x} = - \frac{1}{3}$.

Feb 22, 2015

We can easily find the limit by dividing every term by $x$ as follows
Remember when we put x into the radical we must square it
I am assuming positive infinity. In other words x greater than or equal to zero.

$\frac{\sqrt{{x}^{2} / {x}^{2} + \frac{1}{x} ^ 2}}{\frac{3 x}{x} - \frac{1}{x}}$

$\frac{\sqrt{1 + \frac{1}{x} ^ 2}}{3 - \frac{1}{x}}$

Now taking the limit as x approaches infinity we have

$\frac{\sqrt{1 + 0}}{3 - 0} = \frac{\sqrt{1}}{3} = \frac{1}{3}$