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 {\frac {d}{dx}\sin (x)}{\frac {d}{dx} 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 {\frac {d}{dx}\left (x^2+1\right )}{\frac {d}{dx}\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.
2025-01-06 20:05:06