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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?
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.
$$\lim_{x\to\infty}\left({x-1\over x + 4}\right)^{3x + 2}= \lim_{x\to\infty} \left(1 - {5\over x +4}\right)^{3x} \left(1 - {5\over x +4}\right)^2$$
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 $e^{-15}.$
