5
$\begingroup$

I can't evaluate indeterminate form of a limit like this: $$\lim \limits_{x\to \infty} \left (\frac {x-1}{x+4}\right)^{3x+2}$$ I tried to solve this problem by multiplying fractions' top and bottom by the conjugate of the denominator. I did it many times but I don't have any success and I even don't know if this way right or wrong.

How this limit can be solved?

$\endgroup$
  • $\begingroup$ Do you know that $\left (\dfrac {x-1}{x+4}\right)^{3x+2}$ is short for something else? $\endgroup$ – Git Gud Oct 12 '13 at 21:44
  • 1
    $\begingroup$ Hint: ${x-1\over x+4}=1-{5\over x+4}$. $\endgroup$ – David Mitra Oct 12 '13 at 21:47
  • 1
    $\begingroup$ Thank you very much guys! You helped me a lot! $\endgroup$ – k1ber Oct 12 '13 at 21:52
6
$\begingroup$

It will be easier to rewrite this limit in the form: $$ \lim_{x} \left( \frac{1-1/x}{1+4/x} \right)^{3x+2}. $$ Now, it will suffice to learn how to compute limits of the form: $$ \lim_{x} (1+a/x)^x.$$ This is not very difficult. You can show, for example, that $\ln(1+a/x) = a/x + O(1/x^2)$, so $1+a/x = e^{a/x + O(1/x^2)}$, and finally $(1+a/x)^x = e^{a+O(1/x)}$. Therefore, $\lim_{x} (1+a/x)^x = e^a$. Applying this in the limit you want to compute, we get:

$$\lim_{x} \left( \frac{1-1/x}{1+4/x} \right)^{3x+2} = \left(\frac{e^{-1}}{e^4}\right)^3= 1/e^{15}$$.


Note that the "big O" notation was used. Vaguely speaking, $O(1/x^2)$ stands for a function that tends to $0$ at least as fast as $1/x^2$ does for $x \to \infty$.

$\endgroup$
3
$\begingroup$

Hint : Given limit is of the form $ 1^\infty $, It's a indeterminate form. You can have a look of this to find the limit in such cases.

Why is $1^{\infty}$ considered to be an indeterminate form

$\endgroup$
1
$\begingroup$

$$\lim_{x\to\infty}\left({x-1\over x + 4}\right)^{3x + 2}= \lim_{x\to\infty} \left(1 - {5\over x +4}\right)^{3x} \left(1 - {5\over x +4}\right)^2$$

The second factor goes to 1 at \infty, so what we have left is

$$\lim_{x\to\infty} \left(1 - {5\over x +4}\right)^{3x}.$$ This last limit converges $e^{-15}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.