Question

plzz answitjout l hospital ...limit class 12th​

Answer 1

Answer:

y = lim x->0 [ (1+x)^(1/x) / e ]^1/x

=> logy = limx->0 log[ (1+x)^1/x / e]^1/x

=> logy = lim x->0 [ log( (1+x)^1/x) / e ]/x

=> logy = lim x->0 ( log(1+x)^1/x - loge ]/x

=> logy= lim x->0 [ (1/x)log(1+x) - loge ]/x

=> logy = limx->0 [ log(1+x) - xloge ]/x²

=> logy = limx->0 [ 1/(1+x) - (1)loge ]2x

=> logy = limx->0 [ -1/(1+x)² -(0) ]/2

put limit

=>logy = ( -1/1)/2

=> logy = -1/2

=> y = e^-1/2

Answer 2

just follow the basics...

and also L hospital's rule is used only in the limits of form 0/0 or infinitiy/infinitiy

in which you differentiate number and denominator till the inderminancy vanishes...

you should learn about it ..