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If you substitute 1/2 into the numerator it becomes 2(1/2)^2-5(1/2)-3, or 1/2-5/2-3 =-7. Now do the same substitution in the denominator and you get 6(1/2)^2-7(1/2)+2 = 3/2-7/2+2=0. So the limit s x-> 1/2 = -7/0 = negative infinity.
However, I wonder if perhaps you have a typo in the numerator of this problem? Should it be 2x^2+5x-3 instead of 2x^2-5x-3? If so, then simply substituting x=1/2 yields 0/0 which is undefined, so a different approach is needed. Try factoring both the numerator and denominator and see if there is a common factor that can be elimated. When you do, the limit becomes clear. |
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