# Limits Involving Infinity

## Key Questions

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

#### 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.

$L i {m}_{x \to \infty} x = \infty$

#### Explanation:

Break the problem down into words: "What happens to a function, $x ,$ as we continue increasing $x$ without bound?"

$x$ would also increase without bound, or go to $\infty .$

Graphically, this tells us that as we continue heading right on the $x$-axis (increasing values of $x ,$ going to oo) our function, which is just a line in this case, keeps heading upwards (increasing) with no restrictions.

graph{y=x [-10, 10, -5, 5]}

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

The reason for this is that the domain of the square root function is $x \ge 0$. The limit is undefined if the limit is not being evaluated in the domain.

I usually do this informally.

#### Explanation:

I ask myself what kinds of numbers do I get if I put in more and more negative numbers for $x$. (Often described by saying "bigger and bigger negative numbers".) ("Big" means "far from zero".)

Examples:

Example 1
$f \left(x\right) = 3 {x}^{4} - 7 {x}^{3} + 2 x + 72$

For very very big numbers, the only term that matters is the largest power term: $3 {x}^{4}$. As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?

${x}^{4}$ will always be positive and when I multiply by $3$ the answer will still be positive, so I get bigger and bigger positive numbers.

${\lim}_{x \rightarrow - \infty} f \left(x\right) = \infty$

Example 2
$g \left(x\right) = 5 {x}^{7} + 43 {x}^{4} + 2 {x}^{3} - 5 x + 21$

For very very big numbers, the only term that matters is the largest power term: $5 {x}^{7}$. As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?

${x}^{7}$ will be negative for negative $x$'s and when I multiply by $5$ the answer will still be negative, so I get bigger and bigger negative numbers.

${\lim}_{x \rightarrow - \infty} g \left(x\right) = - \infty$

Example 3 (last)
$h \left(x\right) = - 8 {x}^{6} + 7 x - 3$

For very very big numbers, the only term that matters is the largest power term: $- 8 {x}^{6}$. As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?

${x}^{6}$ will be positive for all $x$'s and when I multiply by $- 8$ the answer will become negative, so I get bigger and bigger negative numbers.

${\lim}_{x \rightarrow - \infty} h \left(x\right) = - \infty$