IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

hsbhatt

Find lim_x-> x²/e^x

johri_anshuman

e^x increases more than x^2 when x>0

this can be shown by taking derivatives

e^x => e^x
and
x^2 => 2x

and e^x attains greater values than x for x>0

so e^x increases more steeply than x^2

so if x-->infinity

e^x >> x^2

=>the reqd limit is 0

raulrag009

use l-hospital rule

u will get 0 as the answer

astatine19

By L'Hospital's rule,

Lim x²
x->

e^x

= Lim 2x
x->

e^x=Lim 2
x->

e^x

= 0

hsbhatt

One other way is to see the polynomial expansion of e^x. It has terms of higher degree than 2 with +ve coefficients. So limit will be 0 for x

I put up this prob bcos of this instructive solution I saw somewhere:

Let f(x) = x³e^-x. Then f '(x) = (3x² - x³) e^-x < 0 for x > 3. Hence f(x) < f(3) for x > 3, so x² e^-x < f(3)/x for x > 3. Hence x² e^-x tends to zero.

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