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Limits And Derivatives

Question
CBSEENMA11013122

Evaluate   space space space space space space space space space space limit as straight x rightwards arrow 2 of fraction numerator 3 straight x squared minus straight x minus 10 over denominator straight x squared minus 4 end fraction

Solution

Evaluating the function at x =2, it is of the form 0 over 0.
Therefore,  WiredFaculty
                                                                                          WiredFaculty
              space space space space space space space space space space space equals fraction numerator 3 left parenthesis 2 right parenthesis plus 5 over denominator 2 plus 2 end fraction equals 11 over 4
                                                                                                                                                                                                                                                                                                                                                                                                                                    

Some More Questions From Limits and Derivatives Chapter