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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Here in the above problem,only $c$ has been replaced by $2c?$ Then should the value of $\lim_{x \to 0} \left[\frac {1+2cx}{1-2cx}\right]^{\frac {1}{x}}$ remain same?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.

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does the question say for a specific c? or c can be any real number? – Arjang Feb 23 '13 at 4:15
@arjang: The limit $\displaystyle\lim_{x \to 0} \left( \frac{1 + cx}{1 - cx} \right)^{\frac{1}{x}}$ will depend on $c$, so the value of $c$ must be fixed. – JavaMan Feb 23 '13 at 4:59
up vote 4 down vote accepted

Let $2x=y$. Then $\frac{1}{x}=\frac{2}{y}$. Note that $y\to 0$ as $x\to 0$. Our expression now becomes $$\lim_{y\to 0}\left(\left(\frac{1+cy}{1-cy}\right)^{1/y}\right)^2.$$

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If $$L=\lim_{x \rightarrow 0} \left ( \frac{1+c x}{1-c x} \right )^{1/x}$$


$$\log{L} = \lim_{x \rightarrow 0} \frac{1}{x} \log{\left ( \frac{1+c x}{1-c x} \right )}$$

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.

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Farfetched, but effective. – Pedro Tamaroff Feb 23 '13 at 4:03
@PeterTamaroff: uhhh...thanks. I think. – Ron Gordon Feb 23 '13 at 4:08
Maybe I should say "This is probably the last thing I would've thought of using". – Pedro Tamaroff Feb 23 '13 at 4:15
@PeterTamaroff: Huh. It seemed like the natural way to go to me. Oh well, different strokes, etc. Thanks! – Ron Gordon Feb 23 '13 at 4:21
@Peter: I think finding the value of $c$ that makes the limit equal to $4$ is quite natural. – JavaMan Feb 23 '13 at 4:22

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.

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No, this is not true. $\lim_{x\to 1} kx$ depends on $k$. Just because for some $k$ it is $=4$ it doesn't mean changing the symbol will leave the value unaffected. – Pedro Tamaroff Feb 23 '13 at 4:03
@PeterTamaroff : but that doesn't make sense, k is meant to be any arbitrary constant, not a specific number. Any examples of such behaviour? – Arjang Feb 23 '13 at 4:13
$\lim_{x\to 1 }2x=2$ but $\lim_{x\to 1 }3x=3$, as I told you. See the other answers and you'll see where you went awry. – Pedro Tamaroff Feb 23 '13 at 4:15
It is not an arbitrary constant, since the limit has a specified value. – Pedro Tamaroff Feb 23 '13 at 4:17
The question is just fine as it stands. Just pay a little more attention next time. You have $\lim F(x,c)=4$ and you want to find $\lim F(x,2c)$. As Andre and rlgordonma showed, it can be evaluated both in terms of $\lim F(x,c)$ and by finding $c$ explicitly. – Pedro Tamaroff Feb 23 '13 at 4:29

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}$

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You're starting with the wrong limit. You know that the initial limit is $4$, so you want to start with $F(x,2c)$ and with $2x=y$ transform it to get something in terms of $F(x,c)$. Look at what Andre did. – Pedro Tamaroff Feb 23 '13 at 16:14

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