Question

without applying l'hospital rule what is the answer of lim x->inf e^x / x^2, how come is it inf as the answer of lim x->inf e^x is inf & the answer of lim x->inf 1/x^2 is 0. and the answer of lim x->inf e^x /x^2 will be inf/0, all I want to ask is that can we divide inf by zero ?

Answer 1

as x --> infinity, you get e^infty/infty^2
but e^x grows faster than x^2, so the expression goes to infinity.

Answer 2

that's some scary dp bro..

Answer 3

phi;
how come lim x->inf 1/x^2 is inf? as http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx it says that it is zero

Answer 4

you were asked about
\[ \frac{e^x}{x^2} \]

Answer 5

which is a ratio of two values that approach infinity

Answer 6

or you can think of it as
\[ e^x \cdot \frac{1}{x^2} \]
which you might view as infinity times zero in the limit.

Answer 7

yeah so if we do that like lim x->inf e^x /x^2 = limx-.inf e^x * limx->inf 1/x^2 = inf *0 ohhhh my misstake my question is now that if what is the answer of inf * 0

Answer 8

opppsss sorry i was dividing inf by zero , is inf * 0 = inf ?

Answer 9

no inf*0 is undefined.

Answer 10

to answer the question, treat it as a ratio of two quantities that both go to infinity.

Answer 11

Then notice that an exponential "grows faster" than x^2, so in the limit, it dominates.

Answer 12

OKKKKKKKKKK thanks

Answer 13

we could use the series definition of e^x to show this.

Answer 14

e^x = 1+x + x^2/2! + x^3/3! +

divide each term by x^2
and take the limit.

Answer 15

just for the practice :) and prolly wrong to begin with
\[\ lim\frac{e^x}{x^2}\frac{(x-1)^2}{e^{x-1}}\]
\[\ lim\frac{e}{x^2}\frac{(x-1)^2}{1}\]
\[\ lim\frac{e}{x^2}\frac{x^2-2x+1}{1}\]
\[\ lim\ e\left(\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{1}{x^2}\right)\]
\[\ lim\ e\]

hmm

Answer 16

thanks amistre

Answer 17

yep, i just cant recall what the result is spose to tell us :)