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Limits And Derivatives

Question
CBSEENMA11013120

Evaluate  space space space space space space space space space space space space space space stack space lim with straight x rightwards arrow 3 below fraction numerator 3 straight x squared minus straight x minus 10 over denominator straight x squared minus 4 end fraction.

Solution

Evaluating the funtion at x =2, it is of the form WiredFaculty.
Therefore,  space space space space space 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 to the power of 2 straight x end exponent minus 4 end fraction equals limit as straight x rightwards arrow 2 of fraction numerator 3 straight x squared minus 6 straight x plus 5 straight x plus 10 over denominator straight x squared minus 2 squared end fraction equals limit as straight z rightwards arrow 2 of fraction numerator left parenthesis straight x minus 2 right parenthesis left parenthesis 3 straight x plus 5 right parenthesis over denominator left parenthesis straight x minus 2 right parenthesis left parenthesis straight x plus 2 right parenthesis end fraction equals limit as straight x rightwards arrow 2 of fraction numerator 3 straight x plus 5 over denominator straight x plus 2 end fraction
                                                        
                                                                                     space space space space space space space space space open square brackets because space straight x space not equal to 2 rightwards double arrow straight x minus 2 not identical to 0 close square brackets
                space space space space space space space space equals space space fraction numerator 2 left parenthesis 2 minus 2 right parenthesis over denominator 2 plus 2 end fraction equals fraction numerator 2 left parenthesis 0 right parenthesis over denominator 4 end fraction equals 0

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