In general how would i simplify these expressions showing enough proof? and is there a website where i see all the rules for limits?
1. Limit Sin ax/Sin bx
x -->0
2. Limit Tan ax/bx
x -->0
3. Limit Cos x/x
x -->0
Here how about this,
Let $\displaystyle f(x)$ and $\displaystyle g(x)$ be two functions such that
$\displaystyle f(x)\sim{g(x)}\quad\text{As }x\to{c}$
Or in other words
$\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=1$
Then you can replace $\displaystyle f(x)$ with $\displaystyle g(x)$ in any non-related limit that goes to c.
Here I will show why
Say we are still talking about $\displaystyle f(x)\sim{g(x)}$
But we need to compute the limit $\displaystyle \lim_{x\to{c}}\frac{f(x)}{h(x)}$
And if we could replace $\displaystyle f(x)$ with $\displaystyle g(x)$ this would be much simpler.
So here is how we can see we can
Since
$\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=1$
From our laws of limits we see that
$\displaystyle \frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}g(x)}=1$
Assuming both limits exist, we then solve for the limit in question
$\displaystyle \lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)$.
Now you may be wondering how this helps, but once again consider that our limit we wish to compute may be rewritten as
$\displaystyle \frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}h(x)}$
Now we see by the equation above that we may rewrite this as
$\displaystyle \frac{\lim_{x\to{c}}g(x)}{\lim_{x\to{c}}h(x)}=\lim _{x\to{c}}\frac{g(x)}{h(x)}$
Which is what we desired.
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So now for the first example what you may do is this
Since
$\displaystyle \lim_{x\to{0}}\frac{\sin(ax)}{ax}=1$
By the substitution $\displaystyle ax=\varphi$
and similarly for $\displaystyle \sin(bx)$
We may say that
$\displaystyle \sin(ax)\sim{ax}$
and
$\displaystyle \sin(bx)\sim{bx}$
So we may rewrite our limit as follows
$\displaystyle \lim_{x\to{0}}\frac{\sin(ax)}{\sin(bx)}\sim\lim_{x \to{0}}\frac{ax}{bx}=\frac{a}{b}$
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Similarly by a substitution of $\displaystyle ax=\varphi$ you can show that
$\displaystyle \tan(ax)\sim{ax}\quad\text{As }x\to{0}$
Or just consider that $\displaystyle \tan(ax)=\frac{\sin(ax)}{\cos(ax)}$
and as $\displaystyle x\to{0}\Rightarrow\cos(ax)\to{1}$
So we can see that
$\displaystyle \lim_{x\to{0}}\frac{\tan(ax)}{bx}\sim\lim_{x\to{0} }\frac{ax}{bx}=\frac{a}{b}$
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The last limit does not exist, to make it a little more clear we can see that
$\displaystyle \lim_{x\to{0}}\frac{\cos(x)}{1}=1$
since $\displaystyle \cos0)=1$
So $\displaystyle \cos(x)\sim{1}\quad\text{As }x\to{0}$
So we may rewrite this as
$\displaystyle \lim_{x\to{0}}\frac{\cos(x)}{x}\sim\lim_{x\to{0}}\ frac{1}{x}$
In which case it should be apparent the limit does not exist since
$\displaystyle \lim_{x\to{0^-}}\frac{1}{x}\ne\lim_{x\to{0^+}}\frac{1}{x}$
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Note the first two limits could have also been done as follows
$\displaystyle \lim_{x\to{0}}\frac{\sin(ax)}{\sin(bx)}=\lim_{x\to {0}}\frac{ax}{bx}\cdot\frac{\sin(ax)}{ax}\cdot\fra c{bx}{\sin(bx)}$
Now rewriting this limit as the products of its individual components we see that we have
$\displaystyle \frac{a}{b}\cdot\lim_{x\to{0}}\frac{\sin(ax)}{ax}\ cdot\lim_{x\to{0}}\frac{bx}{\sin(bx)}$
The last two limits can be found by making the sbustitutions $\displaystyle \varphi=ax$ and $\displaystyle \psi=bx$ respectively.
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and also we could rewrite
$\displaystyle \lim_{x\to{0}}\frac{\tan(ax)}{bx}=\lim_{x\to{0}}=\ lim_{x\to{0}}\frac{\sin(ax)}{\cos(ax)bx}$
Now rewriting it as
$\displaystyle \lim_{x\to{0}}\frac{\sin(ax)}{bx}\cdot\lim_{x\to{c }}\frac{1}{\cos(ax)}=\lim_{x\to{0}}\frac{\sin(ax)} {bx}=\frac{a}{b}\lim_{x\to{0}}\frac{\sin(ax)}{ax}$
where the aforementioned substitution should be made.