8.3: Infinite Limits
GeeksRUs sells titanium mechanical pencils to computer algorithm designers. In an effort to attract more business, they decide to run a rather unusual promotion:
"SALE!! The more you buy, the more you save! Pencils are now \begin{align*}$ \frac{12x}{x  3}\end{align*}
If the zillionaire Spug Dense comes in and says he wants to buy as many pencils as GeeksRUs can turn out, what will the cost of the pencils approach as the order gets bigger and bigger?
Watch This
Embedded Video:
 James Sousa: Limits at Infinity
Guidance
Sometimes, a function may not be defined at a particular number, but as values are input closer and closer to the undefined number, a limit on the output may not exist. For example, for the function f(x) = 1/x (shown in the figures below), as x values are taken closer and closer to 0 from the right, the function increases indefinitely. Also, as x values are taken closer and closer to 0 from the left, the function decreases indefinitely.
We describe these limiting behaviors by writing


\begin{align*}\lim_{x \rightarrow 0^+} \frac{1} {x} = + \infty\end{align*}
limx→0+1x=+∞

\begin{align*}\lim_{x \rightarrow 0^+} \frac{1} {x} = + \infty\end{align*}


\begin{align*}\lim_{x \rightarrow 0^} \frac{1} {x} =  \infty\end{align*}
limx→0−1x=−∞

\begin{align*}\lim_{x \rightarrow 0^} \frac{1} {x} =  \infty\end{align*}
Sometimes we want to know the behavior of f(x) as x increases or decreases without bound. In this case we are interested in the end behavior of the function, a concept you have likely explored before. For example, what is the value of f(x) = 1/x as x increases or decreases without bound? That is,


\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = ?\end{align*}
limx→+∞1x=?

\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = ?\end{align*}


\begin{align*}\lim_{x \to \infty} \frac{1} {x} = ?\end{align*}
limx→−∞1x=?

\begin{align*}\lim_{x \to \infty} \frac{1} {x} = ?\end{align*}
As you can see from the graphs (shown below), as x decreases without bound, the values of f(x) = 1/x are negative and get closer and closer to 0. On the other hand, as x increases without bound, the values of f(x) = 1/x are positive and still get closer and closer to 0.
That is,


\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = 0\end{align*}
limx→+∞1x=0 
\begin{align*}\lim_{x \to \infty} \frac{1} {x} = 0\end{align*}
limx→−∞1x=0

\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = 0\end{align*}
Example A
Evaluate the limit by making a graph:
\begin{align*}\lim_{x\to 3^+} \frac{x + 6}{x  3}\end{align*}
Solution
By looking at the graph:
We can see that as x gets closer and closer to 3 from the positive side, the output increases right out the top of the image, on its way to \begin{align*}\infty\end{align*}
Example B
Evaluate the limit \begin{align*}\lim_{x\to \infty} \frac{11x^3  14x^2 +8x +16} {9x  3}\end{align*}
Solution
To evaluate polynomial function limits, a little bit of intuition helps. Let's think this one through.
First, note that since we are looking at what happens as \begin{align*}x \to \infty\end{align*}
On the top part of the fraction, as x gets truly massive, the 11x^{3} part will get bigger much faster than either of the other terms. In fact, it increases so much faster that the other terms completely cease to matter at all once x gets really monstrous. That means that the important part of the top of the fraction is just the 11x^{3}.
On the bottom, a similar situation develops. As x gets really, really big, the 3 matters less and less. So the bottom may as well be just 9x.
That gives us \begin{align*}\frac{11x^3}{9x}\end{align*}
 Now we can more easily see what happens at the "ends." As x gets bigger and bigger, the numerator continues to get bigger faster than the denominator, so the overall output also increases.
\begin{align*}\therefore \lim_{x\to +\infty} \frac{11x^3  14x^2 +8x +16} {9x  3} = +\infty\end{align*}
Example C
Evaluate \begin{align*}\lim_{x\rightarrow 0} \frac{x + 2} {x + 3} \end{align*}
Solution
This one is easier than it looks! As x > 0, it disappears, leaving just the fraction: 2/3
Concept question wrapup As Spug buys more and more pencils, the cost of each dozen will drop quickly at first, and level out after a while, approaching $12 per dozen. You can see the effect on the graph here: 

Vocabulary
A function with an infinite limit continues to output greater and greater +/ values.
Guided Practice
Questions
1) Make a graph to evaluate the limit \begin{align*}\lim_{x\rightarrow \infty} \frac {1} {\sqrt x} \end{align*}
2) From problem 1, evaluate \begin{align*}\lim_{x\rightarrow 0^{+}} \frac {1} {\sqrt x} \end{align*}
3) Graph and evaluate the limit:
 \begin{align*}\lim_{x\to 2^{+}} \frac {1} {x  2}\end{align*}
Solutions
1) By looking at the image, we see that as x gets huge, so does \begin{align*}\sqrt{x}\end{align*} which means that 1 is being divided by an everlarger number, and the result is getting smaller and smaller.
 The limit is 0
2) On the same image, we can see that as x gets closer and closer to zero, so does \begin{align*}\sqrt{x}\end{align*} which means that 1 is being divided by an ever smaller number, and the result gets bigger and bigger.
 The limit is \begin{align*}+\infty\end{align*}
3) By looking at the image, we can see that as x gets closer and closer to 2 from the positive direction, 1 gets divided by smaller and smaller numbers, so the result gets larger and larger.
Practice
Evaluate the limits, you may graph if you wish:
 \begin{align*}\lim_{x\to 3^{}} \frac {1} {x  3}\end{align*}
 \begin{align*}\lim_{x\to 4^{+}} \frac {1} {x + 4}\end{align*}
 \begin{align*}\lim_{x\to \left(\frac{8}{3}\right)^{+}} \frac {1} {3x + 8}\end{align*}
 \begin{align*}\lim_{x\to 5^{+}} \frac{\left(x^2+11x+30\right)}{x+5}\end{align*}
 \begin{align*}\lim_{x\to \infty} \frac{\left(x^2+11x+30\right)}{x+5}\end{align*}
Evaluate the limits
 \begin{align*}\lim{x \to \infty}\end{align*} \begin{align*}\frac{11x^3 + 20x^2 + 15x  17}{9x^3 + 5x^2  x  17}=\end{align*}
 \begin{align*}\lim{x \to \infty} 13 =\end{align*}
 \begin{align*}\lim{x \to \infty} \frac{2x + 18}{17x  3}=\end{align*}
 \begin{align*}\lim{x \to \infty} 15=\end{align*}
 \begin{align*}\lim{x \to \infty} 5x^2 + 5x + 14=\end{align*}
 \begin{align*}\lim{x \to \infty}7x + 12=\end{align*}
 \begin{align*}\lim{x \to \infty} 3x + 13=\end{align*}
 \begin{align*}\lim{x \to \infty} \frac{13x  8}{19x^3  11x^2 + x + 4}=\end{align*}
 \begin{align*}\lim{x \to \infty} 17x + 14=\end{align*}
 \begin{align*}\lim{x \to \infty}7x^2  2x  13 =\end{align*}