Solving a Limit: Evaluating limx->0 (e^x - 1- x - (x^2/2))/x^3

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Homework Help Overview

The problem involves evaluating the limit as x approaches 0 for the expression (e^x - 1 - x - (x^2/2))/x^3, which falls under the subject area of calculus, specifically limits and series expansions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for evaluating the limit, including l'Hôpital's rule and Taylor series expansion. There are questions about the application of l'Hôpital's rule and the implications of using it multiple times. Some participants express uncertainty about the derivatives involved and the resulting expressions.

Discussion Status

The discussion is active, with participants sharing their attempts and results. There is a mix of approaches being explored, and while some guidance has been offered regarding the use of l'Hôpital's rule, there is no explicit consensus on the final outcome.

Contextual Notes

Participants are navigating through the complexities of the limit and the derivatives involved, with some questioning the correctness of their calculations and the assumptions made during the evaluation process.

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Homework Statement



Evaluate limx->0 (e^x - 1- x - (x^2/2))/x^3

The Attempt at a Solution



I can't remember how to solve this limit. Do I need to evaluate each part separately? I plugged in the 0 to find that the limit does exist. I just can't seem to figure out what to do next.
 
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Why not use l'Hôpital's rule?
 
Or use the Taylor series of the exponential function.
 
If I use l'Hopital's rule won't I end up with 3x^2 at the bottom?
 
You will want to use l'Hopital multiple times.
 
The limit of a sum is the sum of limits - Lim[a+b]=Lim[a]+Lim
Then as everyone before has said, you're going to want to use l'hospital's rule for the last term
 
I used L'Hopital's rule three times and ended up with limx->0(e^x)/5

Is this correct? It still gives me 0 on top.
 
Here is how I did it in case it might help

lim->0 (e^x)' - 1' - x' - (x^2/2)' /x^3'

= lim->0 (e^x' - 1' - x')/ 3x^2'

= lim->0 (e^x' - 1') / 5x'

= lim x->0 (e^x) /5
 
Very close, but look at your second to thrid line in the denominator.

What is: $$\frac{d}{dx}(3x^{2})$$

After you fix that, evaluate it at 0.

<br /> \lim_{x\to 0}~ e^{x}<br /> should not be 0.
 
Last edited:
  • #10
Thanks!
It should be 6x right?

So is 1/6 the correct answer?
 
  • #11
1/6 is the correct result.
 

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