4

I need help understanding this limit question. Wolfram states the answer as −∞. We aren't allowed to use L'Hopital's. A step by step solution would be greatly appreciated.

$$\lim_{y\to 1^-} \frac{\sqrt {1-y^2}}{y-1} = −∞$$

Em.
  • 16,196
Samin
  • 41
  • 4
    +1 to your instructor for "aren't allowed to use L'Hopital". – Ethan Bolker May 10 '16 at 19:14
  • @EthanBolker Respectfully, I disagree. L'hopital is a powerful tool for evaluating limits, even if it does not provide any insight to the problem. While one should learn other methods of evaluating limits (series, conjugates, etc.), L'Hopital does provide a very quick method for calculating complex limits. – zz20s May 10 '16 at 20:14
  • @zz20s Yes. L'Hopital is a powerful tool. The problem I have with it is that students think of it as an algorithm to invoke always automatically. Most of the time you can learn more about what's really going on with simple algebra (as in this question) or by writing out the first few terms of the power series. – Ethan Bolker May 11 '16 at 00:16
  • L'Hopital is overrated in my opinion. See my rant here: http://math.stackexchange.com/questions/1286699/whats-wrong-with-lhopitals-rule/1286806#1286806 – zhw. May 12 '16 at 16:59

4 Answers4

7

Write $1-y^2 = (1-y)(1+y)$ and simplify

Ant
  • 21,522
5

For $0<y<1$ we have $$\frac{\sqrt{1-y^2}}{y-1} =\frac{\sqrt{(1-y)(1+y)}}{y-1}=-\frac{\sqrt{1-y}\sqrt{1+y}}{\sqrt{1-y}\sqrt{1-y}}=-\sqrt{\frac{1+y}{1-y}}$$ and $\frac{1+y}{1-y}\to+\infty$ as $y\to 1^-$.

Hagen von Eitzen
  • 382,479
0

OK, we can go a little nuts here. How about $arcsiny=t$ Then $t$ goes to the value of $\pi/2$ (From only one side, but as it turns out, it doesn't matter). Your limit expression is now of the form $\frac{cost}{sint-1}$ Multiply top and bottom by $sint+1$ creates a $-cos^2t$ in the denominator through the Pythagorean Theorem. And so the resulting expression becomes $\frac{sint+1}{-cost}$ which does not exist for $\pi/2$. It's trig which is really not needed here, but sometimes thinking outside the box ain't bad either...

imranfat
  • 10,212
0

Avoid being confused by signs and make the substitution $y-1=-t$; then the limit becomes $$ \lim_{t\to0^+}\frac{\sqrt{1-(1-t)^2}}{-t}= \lim_{t\to0^+}-\sqrt{\frac{2t-t^2}{t^2}}= \lim_{t\to0^+}-\sqrt{\frac{2}{t}-1} $$

egreg
  • 244,946