# l'Hospital Rule: True/False

I am currently doing this dodgy online practice quiz for Calculus and no matter what I choose, I keep getting it wrong, can someone tell me which one I made wrong? I am pretty sure they are all right, but I guess the computer doesn't think so..

Alright it seems that the first one is actually false so, the final answer is

F F T T T T F

• The first one is false, since the left-most rational expression is $\frac{(x-1)(2x+1)}{(x-1)(3x-2)},$ so the limit is $\frac{(2)(1)\; + \; 1}{(3)(1) \; - \; 2} = 3.$ See my answer to Finding limit of a quotient to see why you don't have to be good at factoring in order to successfully factor in this kind of situation. – Dave L. Renfro Nov 11 '14 at 20:11
• Shouldn't we take the derivative ? – Belphegor Nov 11 '14 at 20:11
• Oh, I see where the problem is in the first one. The limit is $x \rightarrow 1,$ and the second expression is not an indeterminant form for this limit, so L'Hopital's rule doesn't apply. – Dave L. Renfro Nov 11 '14 at 20:15

hint: we have $\frac{2x^2-x-1}{3x^2-5x+2}=\frac{(2x+1)(x-1)}{(3x-2)(x-1)}$ in the term $\frac{4x-1}{6x-5}$ we can plugg $x=1$ because the term defined for $x=1$
Using L'Hospital's Rule $$\lim_{x\to 1} \frac{2x^2-x-1}{3x^2-5x+2}= \frac{2-1-1}{3-5+2} =\frac{0}{0}$$ So now we have $$\lim_{x\to 1} \frac{\frac{d}{dx}[2x^2-x-1]}{\frac{d}{dx}[3x^2-5x+2]} = \lim_{x\to 1} \frac{4x-1}{6x-5} = \frac{4-1}{6-5} =3$$ I assume that this exercise wanted you to practice L'Hospital's rule for this limit.