Where am i wrong in this limits question?

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Homework Help Overview

The problem involves evaluating the limit as x approaches 0 for the expression (1+x+x^2-e^x)/x^2. The original poster believes their approach leads to an incorrect conclusion, while another participant suggests that a more thorough examination of the terms in the expansion of e^x is necessary.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the limit by breaking it into parts and applying known limits, but they question where their reasoning fails. Other participants challenge the validity of substituting (1-e^x)/x with 1, indicating that this is not a valid identity and suggesting that higher-order terms must be considered.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's reasoning. There is a focus on clarifying the assumptions made regarding the limit and the expansion of e^x, but no consensus has been reached on the correct approach or interpretation of the limit.

Contextual Notes

Participants note that the original poster's approach may overlook important higher-order terms in the Taylor series expansion of e^x, which could affect the limit's evaluation. There is also a distinction made between identities and limits, emphasizing the need for careful consideration of the expressions involved.

vkash
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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
 
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=>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
 
Uncle_John said:
=>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
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.
 
vkash said:
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

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
 
vkash said:
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.

(1-e^x)/x =1 is not an identity. an identity is an algebraic equation that is true for any x belonging to a certain domain. like (1+x)^2 = x^2 + 2x + 1 which is valid for any x in real numbers. but (1-e^x)/x = 1 is only true when x approaches zero. It's not an identity. as Ray Vickson said, you need to consider more terms of the e^x series.
 

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