Limit at Infinity Problems and Solutions
Limits at Infinity: Problems and Solutions
Are you working to solve problems about $\displaystyle{\lim_{x \to \infty}}$ and $\displaystyle{\lim_{x \to\, -\infty}}$? Let’s look at common limit at infinity problems and solutions so you can learn to solve them routinely.
(b) Find $\displaystyle{\lim_{x \to -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right)}.$
\[ \begin{align*}
\lim_{x \to \infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to \infty}3x^3 \\[8px] &= \infty \quad \cmark
\end{align*} \] Your solution can be that quick: you look at the polynomial and immediately know what the answer is based on that largest term.
How do we know $\displaystyle{\lim_{x \to \infty}3x^3 = \infty?}$ The rule is $\displaystyle{\lim_{x \to \infty}x^n = \infty }$ for $n > 0.$ But really you should picture in your head $y = x^3$: as x grows and Grows forever in the positive x-direction, $y = x^3$ grows and Grows forever in the positive y-direction, and so we say it has limit of infinity.
\[ \begin{align*}
\lim_{x \to \infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to \infty}x^3 \left(3 + \frac{947x^2}{x^3} – \frac{\sqrt{x}}{x^3} \right) \\[8px] &= \lim_{x \to \infty}x^3 \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right)
\end{align*} \] Now recall that $\displaystyle{\lim_{x \to \infty}\frac{1}{x^n} = 0}$ for $n > 0,$ and so $\displaystyle{\lim_{x \to \infty}\frac{947}{x} = 0}$ and $\displaystyle{\lim_{x \to \infty}\frac{1}{x^{5/2}} = 0}$. Hence
$$\lim_{x \to \infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) = 3$$
And then
\[ \begin{align*}
\lim_{x \to \infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \left(\lim_{x \to \infty}x^3 \right)\lim_{x \to \infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) \\[8px] &= \left(\lim_{x \to \infty}x^3 \right) (3) \\[8px] &= \infty \quad \cmark
\end{align*} \]
\[ \begin{align*}
\lim_{x \to\, -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to\, -\infty}3x^3 \\[8px] &= -\infty \quad \cmark
\end{align*} \] Your solution can be that quick: you look at the polynomial and immediately know what the answer is based on that largest term.
How do we know $\displaystyle{\lim_{x \to\, -\infty}3x^3 = -\infty?}$ The rule is $\displaystyle{\lim_{x \to\, -\infty}x^n = -\infty }$ for odd $n > 0.$ But really you should picture in your head $y = x^3$: as x grows and Grows forever in the negative x-direction, $y = x^3$ grows and Grows forever in the negative y-direction, and so we say it has limit of negative infinity.
\[ \begin{align*}
\lim_{x \to\, -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to\, -\infty}x^3 \left(3 + \frac{947x^2}{x^3} – \frac{\sqrt{x}}{x^3} \right) \\[8px] &= \lim_{x \to\, -\infty}x^3 \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right)
\end{align*} \] Now recall that $\displaystyle{\lim_{x \to\, -\infty}\frac{1}{x^n} = 0}$ for $n > 0,$ and so $\displaystyle{\lim_{x \to\, -\infty}\frac{947}{x} = 0}$ and $\displaystyle{\lim_{x \to\, -\infty}\frac{1}{x^{5/2}} = 0}$. Hence
$$\lim_{x \to\, -\infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) = 3$$
And then
\[ \begin{align*}
\lim_{x \to\, -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \left(\lim_{x \to\, -\infty}x^3 \right)\lim_{x \to\, -\infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) \\[8px] &= \left(\lim_{x \to\, -\infty}x^3 \right) (3) \\[8px] &= -\infty \quad \cmark
\end{align*} \]
\[ \begin{align*}
\lim_{x \to \infty}\frac{5x^2 -7}{3x^2 + 8} &= \lim_{x \to \infty}\frac{5x^2}{3x^2} \\[8px] &= \frac{5}{3} \quad \cmark
\end{align*} \]
Conceptually, the numerator and denominator are growing at the same rate, as modified only by the coefficients of those largest terms.
By the way, the graph shows that the line $y = \dfrac{5}{3}$ is a horizontal asymptote for this function: the function’s curve gets arbitrarily close to that line as $x \to \infty$.
Here, the largest power in the denominator is $x^2$, so we divide each and every term by $x^2$:
\[ \begin{align*}
\lim_{x \to \infty}\frac{5x^2 -7}{3x^2 + 8} &= \lim_{x \to \infty}\frac{\dfrac{5x^2}{x^2} -\dfrac{7}{x^2}}{\dfrac{3x^2}{x^2} + \dfrac{8}{x^2}} \\[8px]
&= \lim_{x \to \infty}\frac{5 -\dfrac{7}{x^2}}{3 + \dfrac{8}{x^2}} \\[8px]
&= \frac{5 -0}{3 + 0} \\[8px]
&= \frac{5}{3} \quad \cmark
\end{align*} \]
Notice that in going from the second to the third line, we made use of the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x^2} = 0}$.
\lim_{x \to \infty}\frac{4x^3 + 2x -24}{x^4 – x^2 + 84 } &= \lim_{x \to \infty}\frac{4x^3}{x^4} \\[8px] &= 0 \quad \cmark
\end{align*} \] because the highest power in the denominator is greater than the highest power in the numerator. (That’s it; the reasoning is that simple.)
Conceptually, the growth in the denominator “wins out” over the growth in the numerator, meaning the denominator grows large at a faster rate than the numerator does, and so the fraction tends toward zero as x grows and Grows and GROWS in the positive direction.
\[ \begin{align*}
\lim_{x \to \infty}\frac{4x^3 + 2x -24}{x^4 – x^2 + 84 } &= \lim_{x \to \infty}\frac{\dfrac{4x^3}{x^4} + \dfrac{2x}{x^4} -\dfrac{24}{x^4}}{\dfrac{x^4}{x^4} – \dfrac{x^2}{x^4} + \dfrac{84}{x^4} } \\[8px] &= \lim_{x \to \infty}\frac{\dfrac{4}{x} + \dfrac{2}{x^3} -\dfrac{24}{x^4}}{1 – \dfrac{1}{x^2} + \dfrac{84}{x^4} } \\[8px] &= \frac{0 + 0 – 0}{1 – 0 +0} \\[8px] &= 0 \quad \cmark
\end{align*} \] Note that to go from the second to the third line, we used the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}$, and $\displaystyle{\lim_{x \to \infty}\frac{1}{x^3} = 0}$, and so forth.
More specifically, we know that the limit is either $\infty$ or $-\infty$. To determine which, we use our usual approach and look at just the term with the highest power in the numerator and the term with the highest power in the denominator:
\[ \begin{align*}
\lim_{x \to \infty}\frac{x^3 +2}{3x^2 + 4} &= \lim_{x \to \infty}\frac{x^3}{3x^2} \\[8px]
&= \lim_{x \to \infty}\frac{x}{3} \\[8px]
&= \infty \quad \cmark
\end{align*} \] 
The second line in the solution shows that the function approaches $\dfrac{x}{3}$ as x grows large, matching what the graph shows. The limit at infinity is (positive) $\infty$ because the function grows in the positive y-direction forever as x grows larger and Larger in the positive direction.
(b) Find $\displaystyle{\lim_{x \to -\infty} \cos(x)}$.
$\displaystyle{\lim_{x \to \infty} \sin(x)}$ does not exist (DNE)$\quad \cmark$
The limit at infinity does not exist because the function continually oscillates between -1 and 1 forever as x grows and Grows. If you were to walk along the function going to the right, you would just keep going up the hills and down the valleys forever, never approaching a single value. Hence the limit at infinity does not exist.
$\displaystyle{\lim_{x \to -\infty} \cos(x)}$ does not exist (DNE)$\quad \cmark$
The limit at infinity does not exist for the same reason $\displaystyle{\lim_{x \to \infty} \sin(x)}$ does not exist: if you were to walk along the function going to the left forever, you would just keep going up the hills and down the valleys between $y = 1$ and $-1,$ never approaching a single value. Hence the limit does not exist.
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Or We can say : lim x→0 (1+(1/x))^(1/x) = ?
Please see the reply below.
Question is : lim x→0 (1/x)^(1/x)
This is a classic L’Hôpital’s Rule question, which this page doesn’t address. But in case it helps, here are the steps you can use: First let $y = (1/x)^{(1/x)}$. Then
\begin{align*}
\ln y &= \ln(1/x)^{(1/x)} \\[8px]
&= (1/x) \ln (1/x) \\[8px]
&= -\frac{\ln (x)}{x}
\end{align*}
where in the second line we made use of the fact that $\ln a^b = b \ln a,$ and in the third line $\ln(1/x) = -\ln x.$
From there, if you take $ \displaystyle{\lim_{x \to \infty }}$ on both sides, on the right you have $\dfrac{\infty}{\infty}$ and so you can apply L’Hôpital’s Rule. You should find \[\displaystyle{\lim_{x \to \infty } \ln y = 0}\]
and so when you undo the natural log, you find that the initial limit = 1.
We hope that helps!