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Evaluate the following limit $$\lim_{x\to0}\frac{e^x-\sum_{k=0}^{n}\frac{x^k}{k!}}{x^{n+1}}$$ where $n\in\mathbb{N}$.

I want to find this limit, if it is possible, without using L'Hopital's rule, derivatives and function series. I want to solve it algebraically. I tried to reduce this limit to some form where I can use other well-known limits at $x\to0$ such as $\frac{\sin x}{x}=1,\frac{e^x-1}{x}=1,\frac{\ln(1+x)}{x}=1,\frac{(1+x)^k-1}{x}=k,(1+x)^\frac{1}{x}=e$. Also, I can use "exponential beats polynomial" rule $x\ln x=0$.
So far I made many attempts. Because there is $e^x$ in the numerator, I tried to apply $\frac{e^x-1}{x}=1$ somewhere. I wrote limit as $$\lim_{x\to0}\frac{\frac{e^x-1}{x}-\frac{\sum_{k=1}^{n}\frac{x^k}{n!}}{x}}{x^n}$$ But, I cannot apply known limit here because I still have $x^n\to0$ in the denominator.
After that, I tried to substitute $t=e^x$, but it didn't help me. Also, I tried to separate these expressions in the numerator to get something similar to known limits, but I could't.
My another attempt was induction. For $n=1$ limit becomes $$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$ But, I still couldn't reduce this fraction to some well-known form.
Is there any way to solve this limit without using derivatives, inequalities or geometrical approach? How we can apply well-known limits here?

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    It is bit stubborn not to use the Taylor series for $e^x$ to evaluate this. =) – Pedro Nov 05 '15 at 19:18
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    @PedroTamaroff. Using Taylor series here makes this limit pretty simple. –  Nov 05 '15 at 19:19
  • On the same lines, can use the Lagrange form of the remainder. – André Nicolas Nov 05 '15 at 19:20
  • What definition of $e^x$ may we use? – Nex Nov 05 '15 at 19:20
  • @Mathematician171 And that's good. We have theorems for such purpose. – Pedro Nov 05 '15 at 19:53
  • While I am myself a fan of the well-known/standard limits technique and I advocate its use over any other technique for evaluating limits, I do understand the limitation of this approach. There are certain limit problems which will require you to use either derivatives (L'Hospital's Rule, Taylor's series) or integation or Squeeze theorem (inequalities). The problem mentioned here requires one of these techniques and is not possible to solve by just using well-known limits. Continued... – Paramanand Singh Nov 06 '15 at 04:24
  • However if one assumes that the limit is question exists (this assumption can be justified only by using the techniques mentioned in previous comment), then it is possible to evaluate it algebraically. See a very good answer http://math.stackexchange.com/a/1182250/72031. – Paramanand Singh Nov 06 '15 at 04:27
  • Another approach would be use $e^{x} = \lim_{n \to \infty}(1 + (x/n))^{n}$ and then use binomial theorem and some theorems in the theory of sequences and series. – Paramanand Singh Nov 06 '15 at 04:31

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By the Taylor Theorem, $$ e^x=\sum_{k=0}^n\frac{x^n}{n!}+\frac{e^\theta}{(n+1)!}x^{n+1} $$ where $\theta$ is between $0$ and $x$. So \begin{eqnarray} \lim_{x\to0}\frac{e^x-\sum_{k=0}^n\frac{x^n}{n!}}{x^{n+1}}=\lim_{x\to0}\frac{e^\theta}{(n+1)!}=\frac{1}{(n+1)!}. \end{eqnarray}

xpaul
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