Evaluate the Limit $\lim_{x\to 0} {\left((e^x - (1+x)) \over x^n\right)}$

Question

Evaluate the Limit:

$$\lim_{x\to 0} {\left((e^x - (1+x)) \over x^n\right)}$$

I am trying to understand how to do this. I have to use series expansion and not L'Hospital. Any help would be great. Thanks! Steve

Answer 1

If you can use L'Hospital, it is straightforward. Remember you must differentiate both numerator and denominator separately. If $n > 2$: $$\lim_{x \to 0}\frac{e^x-(1+x)}{x^n} = \lim_{x \to 0}\frac{e^x-1}{nx^{n-1}} = \lim_{x \to 0}\frac{e^x}{n(n-1)x^{n-2}},$$which gives $+\infty$ coming from the right, and can be $+\infty$ or $-\infty$ from the left, depending on the parity of $n$. If $n = 2$ L'Hospital gives $1/2$ as a result.

If $n = 0$ the result is trivially $0$. If $n = 1$ you can break the limit in $$\lim_{x \to 0}\frac{e^x-1}{x} - 1,$$ which is zero by one of the fundamental results.

Answer 2

Hint: Use series expansion of $e^x$.

Note that $e^x= 1+x + \frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}+\cdots$

Answer 3

L'Hospital's Rule:

$$\lim_{x\to 0} {e^x - (1+x) \over x^n}= \lim_{x\to 0} {e^x - (1+x) \over x^2}.x^{2-n}=\lim_{x\to 0} {(e^x - (1+x))' \over (x^2)'}\lim_{x\to 0} x^{2-n}= $$$$=\lim_{x\to 0} {e^x - 1 \over 2x}\lim_{x\to 0} x^{2-n}= \lim_{x\to 0} {(e^x - 1)' \over 2(x)'}\lim_{x\to 0} x^{2-n}= \frac{1}{2}\lim_{x\to 0} x^{2-n}.$$

Conclusion:

$0$, for $n<2$;

$\frac{1}{2}$ for $n=2$;

$ \infty$ for $n>2$ even;

there is no limit, for $n>2$ odd .

Answer 4

first let us get rid of the easy case of $n \le 0.$ in that case the limit is zero. for the case $n > 0,$ we will make a change of variable $x^n = t, x = t^{1/n}$ so $$\lim_{x\to 0} {\left((e^x - (1+x)) \over x^n\right)} = \lim_{t\to 0} {\left((e^{t^{1/n}} - (1+t^{1/n})) \over t\right)} = \lim_{t\to 0} {\left( t^{2/n} + \cdots\over t\right)} = \lim_{t\to 0} t^{(2-n)/n} $$

so the limit does not exist for $n > 2,$ is $1$ when $n = 2$ and zero for $0 < n < 2$

Answer 5

Using the series expansion, $e^x =1+x+x^2/2+x^3/6+... $. Therefore $e^x-(1+x) =x^2/2+x^3/6+... $ so that $\frac{e^x-(1+x)}{x^2} =1/2+x/6+... $.

From this, $\lim_{x \to 0}\frac{e^x-(1+x)}{x^n} =\begin{cases} 0 & n < 2\\ 1/2 & n=2\\ \infty & n > 2\\ \end{cases} $