Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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Here in the above problem,only $c$ has been replaced by $2c?$ Then should the value of $\lim_{x \to 0} \left[\frac {1+2cx}{1-2cx}\right]^{\frac {1}{x}}$ remain same?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

share|improve this question
    
does the question say for a specific c? or c can be any real number? –  Arjang Feb 23 '13 at 4:15
1  
@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
add comment

4 Answers

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 $$\lim_{y\to 0}\left(\left(\frac{1+cy}{1-cy}\right)^{1/y}\right)^2.$$

share|improve this answer
add comment

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.

share|improve this answer
    
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
show 2 more comments

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.

share|improve this answer
1  
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
2  
$\lim_{x\to 1 }2x=2$ 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
1  
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
show 5 more comments

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}$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.