Calculus/Limit's

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Question
Compute lim x->1 x^2-1/x^2+5x-6 algebraically with factoring and using l'Hopital's Rule

Answer
Note that (x²-1)/(x²+5x-6) = [(x-1)(x+1)]/[x+6)(x-1)].
Cancelling the (x-1) terms on the top and the bottom of the fraction gives (x+1)/(x+6).
Putting in x=1 gives the answer.

Now l'Hospital's Rule says if the numerator and the denominator both go to 0 or both go to infinity, they can each be differentiated.  That is, the derivative of x²-1 is 2x and
the derivative of x²+5x-6 is 2x+5, so the answer can also be found by putting x=1 into
the fraction 2x/(2x+5).

Both of the methods should give 2/7 as the answer.

Calculus

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