Sponsor Area

Relations And Functions

Question
CBSEENMA12036190

The function f: R ~ {0} → R given by
straight f left parenthesis straight x right parenthesis space equals space 1 over straight x minus fraction numerator 2 over denominator straight e to the power of 2 straight x end exponent minus 1 end fraction
can be made continuous at x = 0 by defining f(0) as

  • 2

  • -1

  • 1

  • 0

Solution

C.

1

limit as straight x rightwards arrow 0 of space 1 over straight x space minus fraction numerator 2 over denominator straight e to the power of 2 straight x end exponent minus 1 end fraction

limit as straight x rightwards arrow 0 of space fraction numerator straight e to the power of 2 straight x end exponent minus 1 minus 2 straight x over denominator straight x left parenthesis straight e to the power of 2 straight x end exponent minus 1 right parenthesis end fraction
limit as straight x space rightwards arrow 0 of space fraction numerator 2 straight e to the power of 2 straight x end exponent minus 2 over denominator left parenthesis straight e to the power of 2 straight x end exponent minus 1 right parenthesis plus 2 xe to the power of 2 straight x end exponent end fraction
limit as straight x rightwards arrow 0 of space space fraction numerator 4 straight e to the power of 2 straight x end exponent over denominator 4 straight e to the power of 2 straight x end exponent space plus 4 xe to the power of 2 straight x end exponent end fraction space equals space 1