Start with limits of fractions that you know how to compute. Then take the
derivative of the numerator and the denominator, and see if the new limit equals the
old limit or not.
Two young mathematicians consider a way to compute limits using derivatives.
Check out this dialogue between two calculus students (based on a true story):
- Devyn
- Yo Riley, guess what I did last night?
- Riley
- What?
- Devyn
- I was doing some calculus.
- Riley
- That. Is. Awesome.
- Devyn
- I know! Anyway, I noticed something kinda funny. I think you can sometimes take limits by taking the derivative of the numerator and the denominator.
- Riley
- That’s crazy.
- Devyn
- I know! But check it: \begin{align*} \lim _{x\to 0} \frac {\sin (x)}{x} &= \lim _{x\to 0} \frac {\dd {x}\sin (x)}{\dd {x}x}\\ &= \lim _{x\to 0} \frac {\cos (x)}{1}\\ &=1. \end{align*}
- Riley
- Woah. That. Is. Awes…weird. Hmmm, but it seems like cheating. Wait, it
doesn’t always work, check this out: \[ \lim _{x\to 0} \frac {x^2+1}{x+1} = 1, \]but\begin{align*} \lim _{x\to 0} \frac {\dd {x}\left (x^2+1\right )}{\dd {x}\left (x+1\right )} &= \lim _{x\to 0} \frac {2x}{1} \\ &=0. \end{align*}
Find five examples where this “trick” works, and five examples where it doesn’t
work.