limit of trigonometric functions without L'Hôpital's rule

Question

$\lim_{x\to 0} [\csc^22x-\frac 1 {4x^2}]\ $

and

$\lim_{x\to y} \frac {\tan x- \tan y} {1-\frac{x}{y}(\tan x\tan y+1)}\ $

We have not learn L'Hôpital's rule in my class, we do the $\lim_{x\to 0}\frac{\sin x}{x}=1\ $

I have not came up with anything yet

The second limit does not pose any problem as you can just plug $x=y$ and get $0$ as the answer. The first limit requires the use of advanced tools like Taylor series or L'Hospital's Rule and it won't be possible to get an answer just using algebra of limits. So if you have not studied these advanced techniques then this exercise is misplaced. — Paramanand Singh, Oct 03 '17 at 06:21

alexjo · Answer 1 · 2017-10-02T14:51:17.240

2

HINT Put $u=2x$ so we have $$ \csc^2u-\frac 1 {u^2}=\frac{1}{\sin^2 u}-\frac 1 {u^2}=\frac{ (u+\sin u)(u−\sin u) }{u^2\sin^2u}=\left(1+\frac{\sin u}{u}\right)\frac{ u^2 }{\sin^2u}\frac{u-\sin u}{u^3} $$

Since $\lim_{u\to 0} \frac{\sin u}{u} = \lim_{u\to 0} \frac{u}{\sin u}= 1$ we just need to find $\lim_{u\to 0} \frac{u-\sin u}{u^3}$. This limit is $L=\frac 16$ (see below the link for a proof without derivatives or Taylor or l'Hopital). So the original limit is $$ \lim_{u\to 0}\csc^2u-\frac 1 {u^2}=\lim_{u\to 0}\left(1+\frac{\sin u}{u}\right)\frac{ u^2 }{\sin^2u}\frac{u-\sin u}{u^3}=2\cdot 1\cdot \frac 16=\frac 13 $$

See the answer for $\lim_{u\to 0}\frac{u-\sin u}{u^3}$

For the second limit use the following hint: $$ \frac {\tan x- \tan y} {1-\frac{x}{y}(\tan x\tan y+1)}=\frac{\tan(x-y)[\tan x\tan y+1]}{1-\frac{x}{y}(\tan x\tan y+1)} $$

edited Oct 02 '17 at 14:51

answered Oct 02 '17 at 12:19

alexjo

14,976

it should be minus not plus sorry – Deep J Oct 02 '17 at 12:29
@DeepJ where minus? – alexjo Oct 02 '17 at 12:30
the first problem, i have edited the thread you can see the correct one above – Deep J Oct 02 '17 at 12:33
how do you get $ \ (\frac{2x}{\sin 2x})^2\frac{1}{4x^2}\ $ – Deep J Oct 02 '17 at 12:42
for the second plroblem can I just input x into y ? – Deep J Oct 02 '17 at 12:45

score 0 · Answer 2 · 2017-10-03T09:39:12.133

Let's prove some double-sided estimate for $1/\sin^2 x-1/x^2$ giving the first limit, using only $\lim_{h\to0}\sin h/h=1$ and $\lim_{h\to0}\cos h=1$: $$\frac13\le\frac1{\sin^2 x}-\frac1{x^2}\le\frac1{3\cos^2 x/2}\tag1$$ for all $x\in(0,\pi)$.
Proof: We have $$\sin x=2\sin x/2\,\cos x/2,$$ implying $$\frac1{\sin^2 x}=\frac14\left(\frac1{\sin^2 x/2}+\frac1{\cos^2 x/2}\right)$$ Replacing $x$ by $x/2^k$ and multiplying by $4^{-k}$ gives $$\frac{4^{-k}}{\sin^2 x/2^k}-\frac{4^{-k-1}}{\sin^2 x/2^{k+1}}=\frac{4^{-k-1}}{\cos^2 x/2^{k+1}}$$ and thus (by monotony of $\cos x$) $$4^{-k-1}\le\frac{4^{-k}}{\sin^2 x/2^k}-\frac{4^{-k-1}}{\sin^2 x/2^{k+1}}\le\frac{4^{-k-1}}{\cos^2 x/2}$$ Summing from $k=0$ to $k=n-1$ gives $$\frac13(1-4^{-n})\le\frac1{\sin^2 x}-\frac{4^{-n}}{\sin^2 x/2^n}\le\frac{\frac13(1-4^{-n})}{\cos^2 x/2}$$ Now since $$\frac{4^{-n}}{\sin^2 x/2^n}=\frac1{x^2}\cdot\left(\frac{\sin x/2^n}{x/2^n}\right)^{-2},$$ letting $n\to\infty$ proves our claim (1). qed
It's clear that (1) together with the squeeze lemma means $$\lim_{x\to0}\left(\frac1{\sin^2 x}-\frac1{x^2}\right)=\frac13,$$ and that won't change if we replace $x$ by $2x$.

limit of trigonometric functions without L'Hôpital's rule

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