1. Simplifying limits

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

2. Originally Posted by Misa-Campo
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

Let $f(x)$ and $g(x)$ be two functions such that

$f(x)\sim{g(x)}\quad\text{As }x\to{c}$

Or in other words

$\lim_{x\to{c}}\frac{f(x)}{g(x)}=1$

Then you can replace $f(x)$ with $g(x)$ in any non-related limit that goes to c.

Here I will show why

Say we are still talking about $f(x)\sim{g(x)}$

But we need to compute the limit $\lim_{x\to{c}}\frac{f(x)}{h(x)}$

And if we could replace $f(x)$ with $g(x)$ this would be much simpler.

So here is how we can see we can

Since

$\lim_{x\to{c}}\frac{f(x)}{g(x)}=1$

From our laws of limits we see that

$\frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}g(x)}=1$

Assuming both limits exist, we then solve for the limit in question

$\lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)$.

Now you may be wondering how this helps, but once again consider that our limit we wish to compute may be rewritten as

$\frac{\lim_{x\to{c}}f(x)}{\lim_{x\to{c}}h(x)}$

Now we see by the equation above that we may rewrite this as

$\frac{\lim_{x\to{c}}g(x)}{\lim_{x\to{c}}h(x)}=\lim _{x\to{c}}\frac{g(x)}{h(x)}$

Which is what we desired.
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So now for the first example what you may do is this

Since

$\lim_{x\to{0}}\frac{\sin(ax)}{ax}=1$

By the substitution $ax=\varphi$

and similarly for $\sin(bx)$

We may say that

$\sin(ax)\sim{ax}$

and

$\sin(bx)\sim{bx}$

So we may rewrite our limit as follows

$\lim_{x\to{0}}\frac{\sin(ax)}{\sin(bx)}\sim\lim_{x \to{0}}\frac{ax}{bx}=\frac{a}{b}$
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Similarly by a substitution of $ax=\varphi$ you can show that

$\tan(ax)\sim{ax}\quad\text{As }x\to{0}$

Or just consider that $\tan(ax)=\frac{\sin(ax)}{\cos(ax)}$

and as $x\to{0}\Rightarrow\cos(ax)\to{1}$

So we can see that

$\lim_{x\to{0}}\frac{\tan(ax)}{bx}\sim\lim_{x\to{0} }\frac{ax}{bx}=\frac{a}{b}$
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The last limit does not exist, to make it a little more clear we can see that

$\lim_{x\to{0}}\frac{\cos(x)}{1}=1$

since $\cos0)=1$

So $\cos(x)\sim{1}\quad\text{As }x\to{0}$

So we may rewrite this as

$\lim_{x\to{0}}\frac{\cos(x)}{x}\sim\lim_{x\to{0}}\ frac{1}{x}$

In which case it should be apparent the limit does not exist since

$\lim_{x\to{0^-}}\frac{1}{x}\ne\lim_{x\to{0^+}}\frac{1}{x}$
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Note the first two limits could have also been done as follows

$\lim_{x\to{0}}\frac{\sin(ax)}{\sin(bx)}=\lim_{x\to {0}}\frac{ax}{bx}\cdot\frac{\sin(ax)}{ax}\cdot\fra c{bx}{\sin(bx)}$

Now rewriting this limit as the products of its individual components we see that we have

$\frac{a}{b}\cdot\lim_{x\to{0}}\frac{\sin(ax)}{ax}\ cdot\lim_{x\to{0}}\frac{bx}{\sin(bx)}$

The last two limits can be found by making the sbustitutions $\varphi=ax$ and $\psi=bx$ respectively.
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and also we could rewrite

$\lim_{x\to{0}}\frac{\tan(ax)}{bx}=\lim_{x\to{0}}=\ lim_{x\to{0}}\frac{\sin(ax)}{\cos(ax)bx}$

Now rewriting it as

$\lim_{x\to{0}}\frac{\sin(ax)}{bx}\cdot\lim_{x\to{c }}\frac{1}{\cos(ax)}=\lim_{x\to{0}}\frac{\sin(ax)} {bx}=\frac{a}{b}\lim_{x\to{0}}\frac{\sin(ax)}{ax}$

where the aforementioned substitution should be made.

3. Thanks

Thanks alot man for your time and support, i understand clearly now