# Evaluate $\lim_{x\to0}\frac{e^x-\sum_{k=0}^{n}\frac{x^k}{k!}}{x^{n+1}}$

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?

• It is bit stubborn not to use the Taylor series for $e^x$ to evaluate this. =)
– Pedro
Commented Nov 5, 2015 at 19:18
• @PedroTamaroff. Using Taylor series here makes this limit pretty simple.
– user164524
Commented Nov 5, 2015 at 19:19
• On the same lines, can use the Lagrange form of the remainder. Commented Nov 5, 2015 at 19:20
• What definition of $e^x$ may we use?
– Nex
Commented Nov 5, 2015 at 19:20
• @Mathematician171 And that's good. We have theorems for such purpose.
– Pedro
Commented Nov 5, 2015 at 19:53

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}
• There is a typo, the limit is rather equal to $\frac1{(n+1)!}$. Commented Nov 5, 2015 at 19:26