How can I obtain this division's limit without using derivatives?

Question

$$\lim_{y\to 0} \frac{y}{\cos\left(\frac{\pi}{2}(1+y)\right)}$$ Can anybody help me? I can use basic properties of limits, and some of those basic known limits. I know it would be easier with derivatives, but I was just wondering if it's possible without L-Hospital's rule, derivatives, Taylor series.

Thank you in advance!

My ideas for now: changing cosine into sine. Maybe that. I have no other clue.

score 16 · Answer 1 · edited Jul 31 '15 at 12:38

16

HINT:

$$\cos\left(\dfrac\pi2+A\right)=-\sin A$$

and $$\lim_{h\to0}\dfrac{\sin h}h=\text{ ?}$$

edited Jul 31 '15 at 12:38

Michael Hardy

1

answered Jul 31 '15 at 12:25

lab bhattacharjee

274,582

score 13 · Accepted Answer · edited Jul 31 '15 at 19:03

13

Note that $$\lim_{y\to 0}\frac{y}{\cos\left(\frac{\pi}{2}(1+y)\right)} = \lim_{y\to 0}\frac{y}{\cos\left(\frac{\pi}{2}+\frac{\pi y}{2}\right)} = \lim_{y\to 0}\frac{y}{-\sin\left(\frac{\pi y}{2}\right)} \\ = -\frac{2}{\pi}\lim_{y\to 0}\frac{\left(\frac{\pi y}{2}\right)}{\sin\left(\frac{\pi y}{2}\right)} = -\frac{2}{\pi}(1) = -\frac{2}{\pi}$$

edited Jul 31 '15 at 19:03

einpoklum

359

answered Jul 31 '15 at 12:39

Harish Chandra Rajpoot

37,450

1

Thank you for your help :) – MathIsTheWayOfLife Jul 31 '15 at 12:47
How come we are assuming 0 / 0 is 1, or even that the limit approaches 1. . ? – Andrew Jul 31 '15 at 23:58
@Lemony: We aren't: we're using one of the "basic known limits" involving $\sin$. – Aug 01 '15 at 01:27
5

@Hukyl: absolutely right. – Harish Chandra Rajpoot Aug 01 '15 at 01:30
@HarishChandraRajpoot But how would you derive that limit without Taylor series, derivatives or L'Hôpital's rule? – Kwin van der Veen Aug 01 '15 at 02:28
@fibonatic: You can cheat and use the small angle approximation... but I'm not sure that's terribly rigorous. – Kevin Aug 01 '15 at 03:05
@fibonatic: The question was how to solve the problem with the basic limits, not how to prove the basic limits. (and, IIRC, the usual pedagogy in introductory courses is that the derivatives of the trig functions are proven using the basic limits anyways, rather than the other way around) – Aug 01 '15 at 04:04
1

@user246608: $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ is derived geometrically in this answer. – robjohn Aug 01 '15 at 04:34

How can I obtain this division's limit without using derivatives?

2 Answers2