How do I resolve indeterminate limits?

Please How i do resolve indeterminate limits?
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1 Answer
Jim H
Jan 9, 2017

For this particular problem, factor and simplify, then try again.

Explanation:

Because x=-1x=−1 make both the numerator and denominator polynomials equal to 00, we can be sure that x-(-1) = x+1x−(−1)=x+1 is a factor of both.

Therefore, the quotient can be 'reduced' or 'simplified'

2x^2-x-3 = (x+1)(2x-3)2x2−x−3=(x+1)(2x−3) (check by multiplication)

x^3+2x^2+6x+5 = (x+1)(x^2+x+5)x3+2x2+6x+5=(x+1)(x2+x+5) (check by multiplication)

Therefore,

lim_(xrarr-1)(x^3+2x^2+6x+5)/(2x^2-x-3) = lim_(xrarr-1)(x^2+x+5)/(2x-3)

= -1

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