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

Answer 1

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: $$ \frac {\tan x- \tan y} {1-\frac{x}{y}(\tan x\tan y+1)}=\frac{\tan(x-y)[\tan x\tan y+1]}{1-\frac{x}{y}(\tan x\tan y+1)} $$

Answer 2

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 $2x$.