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.

 Apr 2, 2021

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:

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

 Apr 2, 2021

Thank you @textot ! This was exactly what i needed!

RiemannIntegralzzz  Apr 3, 2021

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