Method for computing limit of a function as $x$ tends to zero

Question

I have a question about computing $$\lim_ {x \to 0} \dfrac{(2/x^3)+(1/x^2)+(1/x)+1}{(1/x^3)+1}.$$ I used a shortcut method of dividing by the highest power but I don't think that I can use this method for limits of the function as $x$ tends to $0$. Is there another way?

Answer 1

Start with the numerator $$\frac{2}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1=\frac{x^3+x^2+x+2}{x^3}$$ Now, the denominator $$\frac{1}{x^3}+1=\frac{1+x^3}{x^3}$$ Now, write the ratio and simplify.

I am sure that you can take from here.

Answer 2

Rewrite it as

$$\lim_{x\to 0}\frac{2+x+x^2+x^3}{x^3+1} = \frac{2}{1}=2$$

Answer 3

We have \begin{align} \lim_{x \to 0} \frac{\frac{2}{x^3} + \frac{1}{x^2} + \frac{1}{x} + 1}{\frac{1}{x^3} + 1} &= \lim_{x \to 0} \frac{x^3}{x^3} \cdot \frac{\frac{2}{x^3} + \frac{1}{x^2} + \frac{1}{x} + 1}{\frac{1}{x^3} + 1} \\ &= \lim_{x \to 0} \frac{2 + x+x^2 + x^3}{1+x^3} \\ &= 2 \end{align}

as each of $x,x^2,x^3$ tend to zero as $x \to 0$.