Compute the limit $\lim_{x\to1}{\frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$

Question

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

What I tried: divided by $x^3$

$$\lim_{x\to1}{\frac{2-\frac{2}{x}+\frac{1}{x^2}-\frac{1}{x^3}}{1-\frac{1}{x}+\frac{3}{x^2}-\frac{3}{x^3}}}$$

Then I plug in $x=1$

$$\frac{2-2+1-1}{1-1+3-3}=\frac{0}{0}$$

This is not correct, where did I make my mistake?

Answer 1

We only evaluate it directly if we do not end up with indeterminate form.

Hint:

$$\lim_{x\to 1}{\frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= \lim_{x\to 1}{\frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$

Answer 2

That's not an effective way to solve since the limit is at $x=1$ and not to $\infty$, indeed in that latter case your method would be fine since we obtain

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

As an effective alternative, in this case, we could use l'Hopital for example.

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Edit for more detail requested

By l'Hopital we obtain

$$\lim_{x\to 1}{\frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}\stackrel{H.R.}=\lim_{x\to 1}{\frac{6x^2-4x+1}{3x^2-2x+3}}=\ldots$$

Can you conclude?

Answer 3

Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.