# How do I find the limit as x approaches infinity of the square root function?

May 27, 2016

${\lim}_{x \to \infty} \sqrt{x} = \infty$

#### Explanation:

Intuitively, as there is no bound to how large we can make $\sqrt{x}$ by increasing $x$, we expect that the limit as $x \to \infty$ of $\sqrt{x}$ would be $\infty$. Indeed, if there were such a bound, say ${x}_{0}$, then we would arrive at a contradiction, as $\sqrt{{x}_{0}^{2} + 1} > \sqrt{{x}_{0}^{2}} = {x}_{0}$.

We can, however, approach the problem in a more rigorous manner.

We say that the limit as $x \to \infty$ of a function $f \left(x\right)$ is $\infty$ (alternately $f \left(x\right) \to \infty$ as $x \to \infty$), denoted ${\lim}_{x \to \infty} f \left(x\right) = \infty$, if, for every integer $N > 0$ there exists an integer $M > 0$ such that $x > M$ implies $f \left(x\right) > N$.

Less formally, that means that for any real value, $f \left(x\right)$ will be greater than that value for large enough $x$.

Our claim is that ${\lim}_{x \to \infty} \sqrt{x} = \infty$. Let's prove it using the above definition.

Take any integer $N > 0$, and let $M = {N}^{2}$. Then, for any $x > M$, we have

$\sqrt{x} > \sqrt{M} = \sqrt{{N}^{2}} = N$

We have shown that for any integer $N > 0$ there exists an integer $M > 0$ such that $x > M$ implies $\sqrt{x} > N$, thereby proving that ${\lim}_{x \to \infty} \sqrt{x} = \infty$.

The above method actually can be used to show that ${x}^{k} \to \infty$ as $x \to \infty$ for any $k > 0$. If we start with an arbitrary $N > 0$ and let $M = {N}^{\frac{1}{k}}$, then for $x > M$ we have ${x}^{k} > {M}^{k} = {\left({N}^{\frac{1}{k}}\right)}^{k} = N$. As $\sqrt{x} = {x}^{\frac{1}{2}}$, the above is just a special case of this.