Compute limit which includes $o$

Question

$$\lim\limits_{x \to 1} \frac{(x-1)^2+o((x-1)^2)}{(x-1)^4+o((x-1)^4)}$$ I really do not understand how to deal with the $o$ in order to compute the limit. My guess to this question is that the answer should be $\infty$. Please help me!

Answer 1

Because you have asymptotic notation, you need to have a limit that goes towards infinity. This is not immediately obvious how to do but since you clearly have a variable going towards $0$ it should be easy to convert it to a limit going to infinity. First, clearly you could restate your problem as:

$$ \lim_{x\rightarrow 0} \frac{x^2 + o(x^2)}{x^4 + o(x^4)} $$

Now we can restate it further as:

$$ \lim_{x\rightarrow 0} \frac{\frac{1}{\frac{1}{x^2}} + o\left(\frac{1}{\frac{1}{x^2}}\right)}{\frac{1}{\frac{1}{x^4}} + o\left(\frac{1}{\frac{1}{x^4}}\right)} $$

Now we can call $\frac{1}{x^2} = X$ and restate as:

$$ \lim_{x\rightarrow \infty}\frac{\frac{1}{x} + o\left(\frac{1}{x}\right)}{\frac{1}{x^2} + o\left(\frac{1}{x^2}\right)} $$

At this point, we choose some functions $f \in o\left(\frac{1}{x}\right)$ and $g\in o\left(\frac{1}{x^2}\right)$:

$$ \lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x^2}}\frac{1 + \frac{f(x)}{\frac{1}{x}}}{1 + \frac{g(x)}{\frac{1}{x^2}}} = \lim_{x\rightarrow \infty} x\frac{1 + \frac{f(x)}{\frac{1}{x}}}{1 + \frac{g(x)}{\frac{1}{x^2}}} $$

We know that $\lim_{x\rightarrow \infty} \frac{f(x)}{\frac{1}{x}} = 0$ and $\lim_{x\rightarrow \infty} \frac{g(x)}{\frac{1}{x^2}} = 0$ (by definition of little-o notation).

Using the multiplication rule, this gives:

$$ \lim_{x\rightarrow \infty} x\frac{1 + \frac{f(x)}{\frac{1}{x}}}{1 + \frac{g(x)}{\frac{1}{x^2}}} = \left(\lim_{x\rightarrow \infty} x\right)\cdot \left(\lim_{x\rightarrow \infty}\frac{1 + \frac{f(x)}{\frac{1}{x}}}{1 + \frac{g(x)}{\frac{1}{x^2}}}\right) = \left(\lim_{x\rightarrow \infty}x\right)\cdot 1 $$

Answer 2

Not entirely rigorous as you really need both limits to exist as real numbers to use $\lim f(x)g(x) = \lim f(x)\lim g(x)$, but the idea is:

$$\require{cancel}\lim\limits_{x \to 1} \frac{(x-1)^2+ o((x-1)^2)}{(x-1)^4+o((x-1)^4)} = \lim\limits_{x \to 1} \frac{(x-1)^2}{(x-1)^4}\frac{1+\cancel{\frac{o((x-1)^2)}{(x-1)^2}}}{1+\cancel{\frac{o((x-1)^4)}{(x-1)^4}}} = \lim\limits_{x \to 1} \frac{(x-1)^2}{(x-1)^4} = \infty$$

If you're a bit more careful you could turn this into a rigorous argument, though.