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I have the following expression:

$\displaystyle\lim _{x\to \infty }\left(\dfrac{x+2}{\:x-6}\right)^\left(\dfrac{x}{\:4}\right)$

I'm studying Calculus I and our lector has shown us ways of transforming such limits to:

$\displaystyle\lim_{x\to \infty }\left(1+\frac{1}{\:x}\right)^x = e$

The way this calculator solves it is not immediately clear to me, is there any other way to find the above limit?

Yadati Kiran
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PowerUser
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4 Answers4

3

HINT

We have that

$$\left(\frac{x+2}{x-6}\right)^{\frac x 4}=\left(\frac{x-6+8}{x-6}\right)^{\frac x 4}=\left(1+\frac{8}{x-6}\right)^{\frac x 4}$$

then we can manipulate further in order to use the standard limit.

Refer to the related

user
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  • I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out... – PowerUser Nov 19 '18 at 13:33
  • @PowerUser Use that $A^n=(A^m)^{\frac n m}$. – user Nov 19 '18 at 13:35
  • I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks! – PowerUser Nov 19 '18 at 13:56
  • @PowerUser You are welcome! Bye – user Nov 19 '18 at 13:58
  • @PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) \log (f(x))}$$ to obtain teh same result by standard limit $$t \to 0 \quad \frac{\log (1+t)}t \to 1$$ – user Nov 19 '18 at 14:00
1

You may proceed as follows:

  • Set $y = x-6$ $$\left(\frac{x+2}{x-6} \right)^{\frac{x}{4}} = \left(1 +\frac{8}{y} \right)^{\frac{y+6}{4}} = \left(1 +\frac{2}{\frac{y}{4}} \right)^{\frac{y}{4}}\cdot \left(1 +\frac{8}{y} \right)^{\frac{3}{2}} \stackrel{y \to \infty}{\longrightarrow}e^2$$
trancelocation
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1

$\displaystyle\lim _{x\to \infty }\left(\dfrac{x+2}{\:x-6}\right)^\left(\dfrac{x}{\:4}\right)=\lim _{x\to \infty }\left(\dfrac{1+\dfrac2x}{1-\dfrac6x}\right)^\left(\dfrac{x}{\:4}\right)=\dfrac{\left(\lim_{x\to\infty}\left(1+\dfrac2x\right)^{x/2}\right)^{1/2}}{\left(\lim_{x\to\infty}\left(1-\dfrac6x\right)^{-x/6}\right)^{-3/2}}=\dfrac{e^{1/2}}{e^{-3/2}}=?$

lab bhattacharjee
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If you divide $x+2$ by $x-6$ the quotient is $1$ and the remainder is $8$, so the limit becomes $$\lim_{x\to\infty}\left(1+\frac{8}{x-6}\right)^{x/4}$$

Now, substitute $x/4$ by $$\frac{x-6}8\cdot\frac{8}{x-6}\cdot\frac x4$$ to get

$$\lim_{x\to\infty}\left[\left(1+\frac{1}{\dfrac{x-6}8}\right)^{\dfrac{x-6}8}\right]^{\dfrac{8x}{4(x-6)}}$$

ajotatxe
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