CBSE Class 11-science Answered

What will be the answer for this question?

Asked by dineshchem108 | 02 May, 2019, 08:55: PM

Expert Answer

begin mathsize 16px style limit as straight x rightwards arrow infinity of space open parentheses fraction numerator straight x plus 7 over denominator straight x plus 2 end fraction close parentheses to the power of straight x plus 4 end exponent equals limit as straight x rightwards arrow infinity of space open parentheses 1 plus fraction numerator 5 over denominator straight x plus 2 end fraction close parentheses to the power of straight x plus 2 plus 2 end exponent Put space fraction numerator 1 over denominator straight x plus 2 end fraction equals straight t straight x rightwards arrow infinity space rightwards double arrow straight t rightwards arrow 0 then space limit as straight x rightwards arrow infinity of space open parentheses 1 plus fraction numerator 5 over denominator straight x plus 2 end fraction close parentheses to the power of straight x plus 2 plus 2 end exponent equals limit as straight t rightwards arrow 0 of space open parentheses 1 plus 5 straight t close parentheses to the power of 1 over straight t plus 2 end exponent equals limit as straight t rightwards arrow 0 of space open parentheses 1 plus 5 straight t close parentheses to the power of fraction numerator 1 over denominator 5 straight t end fraction cross times 5 end exponent space open parentheses 1 plus 5 straight t close parentheses squared equals straight e to the power of 5 end style

Answered by Sneha shidid | 03 May, 2019, 09:43: AM