$\lim_{x \to 0} [\frac {1+2cx}{1-2cx}]^{\frac {1}{x}}=?$ I faced the following problem which says:   

If $\lim_{x \to 0} \left[\frac {1+cx}{1-cx}\right]^{\frac {1}{x}}=4,$ then $\lim_{x \to 0}  \left[\frac {1+2cx}{1-2cx}\right]^{\frac {1}{x}}=?$ 

Here in the above problem,only $c$ has been replaced by $2c?$ Then should the value of $\lim_{x \to 0} \left[\frac {1+2cx}{1-2cx}\right]^{\frac {1}{x}}$ remain same? I can not prove it.
Can someone point me in the right direction?Thanks in advance for your time.
 A: Let $2x=y$. Then $\frac{1}{x}=\frac{2}{y}$. Note that $y\to 0$ as $x\to 0$. Our expression now becomes
$$\lim_{y\to 0}\left(\left(\frac{1+cy}{1-cy}\right)^{1/y}\right)^2.$$
A: If $$L=\lim_{x \rightarrow 0} \left ( \frac{1+c x}{1-c x} \right )^{1/x}$$
then
$$\log{L} = \lim_{x \rightarrow 0} \frac{1}{x} \log{\left ( \frac{1+c x}{1-c x} \right )}$$
For small $x$,
$$\log{\left ( \frac{1+c x}{1-c x} \right )} = 2 c x + O(x^3)$$
Therefore, $\log{L} = 2 c = \log{4} \implies c=\log{2}$.
You should be able to do the rest.
A: This is wrong!
In general $\lim F(x,c) = k \quad x,y,c\in \mathbb R \not \Rightarrow \lim F(x,mc) = k$
, read the comments:
Yes, that is true, value will remain the same (if the original premise is true). Instead of 2c use c', that is let c'=2c. the second limit is exactly the same as first limit with c now being called c', but that is just renaming c to c' in the first limit.
A: Warning : dubious operation, criticisms  welcome:
$\lim_{x \to 0} \left[\frac {1+cx}{1-cx}\right]^{\frac {1}{x}} = \lim_{2x \to 0} \left[\frac {1+2cx}{1-2cx}\right]^{\frac {1}{2x}}= \lim_{2x \to 0} \left( \left[\frac {1+2cx}{1-2cx}\right]^{\frac {1}{x}} \right)^{\frac {1}{2}} = 4^{\frac {1}{2}} =2$ 
In above I have used :
$ \lim_{x \to 0} \equiv \lim_{2x \to 0}$
