Calculus can show us how our intuition can fail us, a common theme in philosophy.
This post is part of the series Connecting Calculus to the Real World.
Calculus can show us how our intuition can fail us, a common theme in philosophy. When we question our intuitions and analyze a problem in depth, we sometimes come up with wildly different answers than we had expected. Our confusion can lead us to think about the problem in different ways, and build a new intuition that will actually lead us to the right answer.
We’ll demonstrate this process on a limit problem.
At first glance, the limit
$\begin{align*} \lim_{x \rightarrow \infty} \sqrt{x^2+x} - \sqrt{x^2} \end{align*}$
looks like something we should be able to tackle with intuition.
If we remove the square roots, the limit comes out to infinity.
$\begin{align*} \lim_{x \rightarrow \infty} (x^2+x) - x^2 &= \lim_{x \rightarrow \infty} x \\[5pt] &= \infty \end{align*}$
The square root of infinity is infinity, so shouldn’t the original limit come out to infinity?
It turns out, this is not the case. Here is what happens if we use formal reasoning to figure the limit out step-by-step:
$\begin{align*} \lim_{x \rightarrow \infty} \sqrt{x^2+x} - \sqrt{x^2} &= \lim_{x \rightarrow \infty} \frac{\left( \sqrt{x^2+x} - \sqrt{x^2} \right) \left( \sqrt{x^2+x} + \sqrt{x^2} \right) }{\sqrt{x^2+x} - \sqrt{x^2}} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{(x^2+x)-x^2}{\sqrt{x^2+x} - \sqrt{x^2}} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{x}{\sqrt{x^2+x} - \sqrt{x^2}} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{x \left( \frac{1}{x} \right)}{\left( \sqrt{x^2+x} - \sqrt{x^2} \right) \left( \frac{1}{x} \right) } \\[5pt] &= \lim_{x \rightarrow \infty} \frac{1}{\frac{\sqrt{x^2+x}}{x} - \frac{\sqrt{x^2}}{x} } \\[5pt] &= \lim_{x \rightarrow \infty} \frac{1}{\frac{\sqrt{x^2+x}}{\sqrt{x^2}} - \frac{\sqrt{x^2}}{\sqrt{x^2}} } \\[5pt] &= \lim_{x \rightarrow \infty} \frac{1}{\sqrt{\frac{x^2+x}{x^2}} - 1} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{x}} - 1} \\[5pt] &= \frac{1}{\sqrt{1+0} - 1} \\[5pt] &= \frac{1}{2} \end{align*}$
Where does that number come from? It’s hard to see in the original limit.
However, there is intuition that can motivate the correct answer. If we allow ourselves to add a constant term inside the limit, then it all makes sense:
$\begin{align*} \lim_{x \rightarrow \infty} \sqrt{x^2+x} - \sqrt{x^2} &= \lim_{x \rightarrow \infty} \sqrt{x^2+x+\frac{1}{4}} - \sqrt{x^2} \\[5pt] &= \lim_{x \rightarrow \infty} \sqrt{\left( x + \frac{1}{2} \right)^2} - \sqrt{x^2} \\[5pt] &= \lim_{x \rightarrow \infty} \left( x+\frac{1}{2} \right) - x \\[5pt] &= \lim_{x \rightarrow \infty} \frac{1}{2} \\[5pt] &= \frac{1}{2} \end{align*}$
Now that we’ve rebuilt our intuition about these kinds of problems, we can tackle similar ones by adding constants within square roots. For example, using our new intuition, we find that
$\begin{align*} \lim_{x \rightarrow \infty} \sqrt{x^2+2x} - \sqrt{x^2+5} &= \lim_{x \rightarrow \infty} \sqrt{x^2+2x+1} - \sqrt{x^2} \\[5pt] &= \lim_{x \rightarrow \infty} \sqrt{(x+1)^2} - \sqrt{x^2} \\[5pt] &= \lim_{x \rightarrow \infty} (x+1)-x \\[5pt] &= \lim_{x \rightarrow \infty} 1 \\[5pt] &= 1. \end{align*}$
Working the problem out the long way, we see that our new intuition is correct!
$\begin{align*} \lim_{x \rightarrow \infty} \sqrt{x^2+2x} - \sqrt{x^2+5} &= \lim_{x \rightarrow \infty} \frac{\left( \sqrt{x^2+2x} - \sqrt{x^2+5} \right) \left( \sqrt{x^2+2x} + \sqrt{x^2+5} \right)}{\sqrt{x^2+2x} + \sqrt{x^2+5}} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{(x^2+2x)-(x^2+5)}{\sqrt{x^2+2x} + \sqrt{x^2+5}} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{2x-5}{\sqrt{x^2+2x} + \sqrt{x^2+5}} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{\left(2x-5\right) \left( \frac{1}{x} \right)}{\left(\sqrt{x^2+2x} + \sqrt{x^2+5}\right) \left( \frac{1}{x} \right)} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{2+\frac{5}{x}}{\frac{\sqrt{x^2+2x}}{\sqrt{x^2}} + \frac{\sqrt{x^2+5}}{\sqrt{x^2}}} \\[5pt] &= \lim_{x \rightarrow \infty} \frac{2+\frac{5}{x}}{\sqrt{\frac{x^2+2x}{x^2}} + \sqrt{\frac{x^2+5}{x^2}} } \\[5pt] &= \lim_{x \rightarrow \infty} \frac{2+\frac{5}{x}}{\sqrt{1+\frac{2}{x}} + \sqrt{1+\frac{5}{x^2}} } \\[5pt] &= \frac{2}{\sqrt{1}+\sqrt{1}} \\[5pt] &= 1 \end{align*}$
This post is part of the series Connecting Calculus to the Real World.