How do I resolve indeterminate limits?

Question

Please How i do resolve indeterminate limits?
thank you ( ◑‿◑)ɔ

1262 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-09T03:29:46

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

Because #x=-1# make both the numerator and denominator polynomials equal to #0#, we can be sure that #x-(-1) = x+1# is a factor of both.

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

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

#x^3+2x^2+6x+5 = (x+1)(x^2+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#