I am stuck on this limits problem: \(\lim_{x\rightarrow 1}\frac{x^3-3x^2+3x-1}{x-1}\)

I could easily factor the numerator into $(x-1)^3$ and solve, but the problem asked me specifically to use L'Hopital's method of evaluating limits. I have tested and found that it is an indeterminate form, 0/0, but I have never done L'Hopital's method before(It's an extracurricular thing). I have evaluated the limit to be 0 through factoring and cancelling, but I need to show this through L'Hopital's.

RiemannIntegralzzz Apr 2, 2021

#1**+2 **

L'Hopital's rule tells us that if the numerator function (let's call that f(x)) and the denominator function (let's call that g(x)) both approach 0 as x approaches n, where n is a constant, and if \(\lim_{x\rightarrow n} (\frac{f'(x)}{g'(x)})\) exists, then \(\lim_{x\rightarrow n} (\frac{f'(x)}{g'(x)})\).

So basically keep taking the derivative of the top and bottom of the fraction until you reach a point where it's not indeterminate or it's undefined (like 4/0).

In this case, you need to use the power rule:

\(\lim_{x\rightarrow 1}\frac{x^3-3x^2+3x-1}{x-1}\\=\lim_{x\rightarrow 1}\frac{3x^2-6x+3}{1}\\ 3(1)^2-6(1)+3=0\)

textot Apr 2, 2021