How to compute $\lim_{x\to 0}\frac{e^x -1-x-\frac{x^2}{2}}{x^3}$ without using L'Hospital rule?

Question

Find the following limit without using L'Hospital rule nor series $$\lim_{x\to 0}\frac{e^x -1-x-\frac{x^2}{2}}{x^3}$$ I tried to divide it into two limits such that one of the be $$\lim_{x\to 0}\frac{e^x-x-1}{x^2}$$ because i know it solution but i could not do that ?

Answer 1

My attempt of the general case :|

$$\lim_{x\rightarrow 0} \frac{e^{x}-\sum\limits^{p-1}_{k=0}\frac{x^{k}}{k!}}{x^{p}}= \frac{1}{p!}. $$

Using the definition $e^{x}= \lim_{n \rightarrow \infty} (1+\frac{x}{n})^n$. we will work the expression $$(1+\frac{x}{n})^n-\sum\limits^{p-1}_{k=0}\frac{x^{k}}{k!}= $$ for $n>p$, expand $(1+\frac{x}{n})^n$ by the binomial $$=\sum_{k=0}^{n}{n \choose k}(\frac{x}{n})^k -\sum\limits^{p-1}_{k=0}\frac{x^{k}}{k!}=\underbrace{\sum_{k=0}^{p-1}{n \choose k}(\frac{x}{n})^k + {n \choose p}(\frac{x}{n})^p+\sum_{k=p+1}^{n}{n \choose k}(\frac{x}{n})^k}_{\sum\limits_{k=0}^{n}{n \choose k}(\frac{x}{n})^k} -\sum\limits^{p-1}_{k=0}\frac{x^{k}}{k!} =$$

$$=\sum_{k=0}^{p-1}x^{k}\underbrace{({n \choose k}\frac{1}{n^{k}}-\frac{1}{k!})}_{\rightarrow 0} +\underbrace{{n \choose p}(\frac{x}{n})^p}_{\rightarrow \frac{x^p}{p!}}+\sum_{k=0}^{n-p-1}{n \choose k+p+1}(x)^{k+p+1}\frac{1}{n^{k+p+1}} =$$ we have $\lim_{n \rightarrow \infty} {n \choose k}\frac{1}{n^{k}} =\lim_{n \rightarrow \infty} \frac{1}{k!} (\frac{n}{n})\cdots (\frac{n-k+1}{n})= \frac{1}{k!}$, (every factor has limit 1) so as we take $n \rightarrow \infty $ in the first sum the resultant is zero, dividing the expression by $x^p$ and applying the limit we have $$ \lim_{x \rightarrow 0} \lim_{n \rightarrow \infty} \frac{1}{p!} +\sum_{k=0}^{n-p-1}{n \choose k+p+1}(x)^{k+1}\frac{1}{n^{k+p+1}} = $$ changing the order of the limits $$ \lim_{n \rightarrow \infty} \lim_{x \rightarrow 0} \frac{1}{p!} +\sum_{k=0}^{n-p-1}{n \choose k+p+1}\underbrace{(x)^{k+1}}_{\rightarrow 0}\frac{1}{n^{k+p+1}} = \frac{1}{p!}$$

so $$\lim_{x\rightarrow 0} \frac{e^{x}-\sum\limits^{p-1}_{k=0}\frac{x^{k}}{k!}}{x^{p}}= \frac{1}{p!}. $$

Answer 2

$e^x$ is a (log-)convex function, hence $e^x\geq 1+x$ holds for any $x\in\mathbb{R}$. By integrating both sides over the interval $(0,z)$ we get $e^z\geq 1+z+\frac{z^2}{2}$ for any $z>0$ and $e^z\leq 1+z+\frac{z^2}{2}$ for any $z<0$. Performing the same trick twice we get $e^{z}\geq 1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}$ for any $z>0$ and the converse inequality for any $z<0$. By squeezing, the wanted limit equals $\frac{1}{6}$.

Answer 3

It's basically the LH's rule, but not the simple application of it: $$\lim_{x \to 0}\frac{e^x-1-x-\frac{x^2}{2}}{x^3}$$ Let's call $e^x-x-\frac{x^2}{2}:=f(x)$, so: $$\lim_{x \to 0}\frac{f(x)-1}{x^3}=\lim_{x \to 0}\frac{f(x)-f(0)}{x^3-0}=\lim_{x \to 0}\frac{f(x)-f(0)}{x^3-0^3}=\lim_{x \to 0}\frac{f(x)-f(0)}{x^3-0^3}\cdot\frac{\frac{1}{x-0}}{\frac{1}{x-0}}=\lim_{x \to 0}\frac{\frac{f(x)-f(0)}{x-0}}{\frac{x^3-0^3}{x-0}}=\frac{\lim\limits_{x \to 0}\frac{f(x)-f(0)}{x-0}}{\lim\limits_{x \to 0}\frac{x^3-0^3}{x-0}}\stackrel{def}{=}\frac{f'(0)}{(x^3)'(0)}=\lim_{x \to 0} \frac{e^x-1-x}{3x^2}$$ And doing it again 2 times you can get: $$\lim_{x \to 0} \frac{e^x}{6}=\frac{1}{6}$$
Or using $\lim\limits_{x \to 0}\frac{e^x-1-x}{x^2}$ : $$f(x):=\left\{ \begin{array}{ll} \frac{e^x-1-x}{x^2} & \mbox{if } x \neq 0 \\ \frac{1}{2} & \mbox{if } x =0 \end{array} \right.$$ Now your limit can be rewritten this way: $$\lim\limits_{x \to 0}\frac{f(x)-f(0)}{x}=\lim\limits_{x \to 0}\frac{f(x)-f(0)}{x-0}\stackrel{def}{=}f'(0)=\lim_{x \to 0} \frac{e^x(x-2)+x+2}{x^3}$$ So: $$A:=\lim_{x \to 0} \frac{e^x(x-2)+x+2}{x^3}$$ $$B:=\lim_{x \to 0}\frac{e^x-1-x-\frac{x^2}{2}}{x^3}$$ From this: $$A+2B=\lim_{x \to 0}\frac{(x-2)e^x +x+2+2\left(e^x-1-x-\frac{x^2}{2}\right)}{x^3}=\lim_{x \to 0}\frac{x\left(e^x-x-1\right)}{x^3}=\lim_{x \to 0} \frac{e^x-1-x}{x^2}=\frac{1}{2}=3\lim_{x \to 0}\frac{e^x-1-x-\frac{x^2}{2}}{x^3}$$ So: $$\lim_{x \to 0}\frac{e^x-1-x-\frac{x^2}{2}}{x^3}=\frac{1}{6}$$

Answer 4

If the limit exists, let it be $L$, by the change of variable $x\to-x$ and taking the average,

$$L=\lim_{x\to0}\frac{e^x-1-x-\dfrac{x^2}2}{x^3}=\lim_{x\to0}\frac{e^{-x}-1+x-\dfrac{x^2}2}{-x^3}=\lim_{x\to0}\frac{\sinh x-x}{x^3}.$$

Now with $x\to3x$, and using $\dfrac{\sinh x}x\to1^*$,

$$L=\lim_{x\to0}\frac{\sinh 3x-3x}{27x^3}=\lim_{x\to0}\frac{3\sinh x+4\sinh^3x-3x}{27x^3}=\frac19L+\frac4{27}.$$

Hence,

$$L=\frac16.$$

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$^*$With the same trick,

$$\lim_{x\to0}\frac{e^x-1}x=\lim_{x\to0}\frac{e^{-x}-1}{-x}=\lim_{x\to0}\frac{\sinh x}x=1.$$