Question

lim x tends to zero x²/e^infinite. ​

Answer 1

Answer:

Put y = (1+x^2)^{e^(-x)}. Hence

ln(y) =e^(-x).ln(1+x^2) =[ ln(1+(x^2)]/e^x, which is an indeterminate form of the type inf/inf and so we apply de l'ho^pital’s rule and ratio of derivatives of the numerator and the denominator is [2x/(1+x^2)]/e^x = 2/[1/x)+x].e^x which approaches 0 as x tends to +inf. since the denominator approaches inf. Therefore limit of ln(y) as x tends to+inf is 0. Hence the required limit of y is e^0 = 1.

Step-by-step explanation: