Homework Help: Where am i wrong in this limits question?

1. Jul 22, 2011

vkash

question is
limx->0 (1+x+x^2-e^x)/x^2
How i did it.
=>limx->0 (1-e^x)/x^2+(x+x^2)/x^2
=>limx->0 ((1-e^x)/x)*1/x+1/x+1
using limx->0(e^x-1)/x=1
=>limx->0 (-1)*1/x+1/x+1
=>limx->0 1
=>Answer=1. (hope you understand my writing way. Where are list of tags for this forum)
I think i do nothing wrong but this is wrong answer. Applying L'Hospital rule it's answer will come out to be 1/2 which is correct answer. Tell me where i am wrong

Last edited: Jul 22, 2011
2. Jul 22, 2011

Uncle_John

=>limx->0 ((1-e^x)/x)*1/x+1/x+1
using limx->0(e^x-1)/x=1
=>limx->0 (-1)*1/x+1/x+1
Here you are wrong

3. Jul 22, 2011

vkash

dear here i change (1-e^x)/x by 1 which is an identity. I use this identity also that lim f(x).g(x)=lim f(x).lim g(x) and lim [f(x)+g(x)]=lim f(x)+lim g(x)
friend try to explain it in better way.

4. Jul 22, 2011

Ray Vickson

Your problem is that you look only at the "first-order" terms in (1 - e^x), so as to conclude that lim{ [(1 - e^x)/x] *(1/x) + 1/x] } = 0 as x --> 0. This is false. When we look at (1-e^x) we need to keep also terms of second order in small x, because those will give the nonzero part of lim { [(1 - e^x)/x] *(1/x) + 1/x] }. In other words: if e^x = 1 + x + r(x), we have (1 - e^x)/x = (-x - r(x) )/x = -1 - r(x)/x, hence (1/x)[(1-e^x)/x] = -1/x - r(x)/x^2, so lim { [(1 - e^x)/x] *(1/x) + 1/x] } = - lim r(x)/x^2. This last limit is not zero.

RGV

5. Jul 22, 2011