# How to evaluate indeterminate form of a limit

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?

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

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 $x \to \infty$.

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

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

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 $e^{-15}.$