4

I want to compute $\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x^3}$ without L'Hospital rule or Taylor series

my second question

there is a part that I don't understand which is

$$\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x^3}$$ $$=\lim_{x\to0}\frac{2e^{-x}(e^{2x}-1)}{2xx^2}-\frac{2}{x^2} $$ as $$=\lim_{x\to0}\frac{(e^{2x}-1)}{2x}=1$$ then $$=\lim_{x\to0}\frac{2e^{-x}(e^{2x}-1)}{2xx^2}-\frac{1}{x^2} $$ = $$\lim_{x\to0}\frac{2e^{-x}}{x^2}-\frac{2}{x^2} $$ which is infinity but this answer is wrong the answer is $\frac{1}{3}$ but why my solution was wrong ? isn't $\lim(a+b)=\lim(a)+\lim(b)$ and $\lim(ab)=\lim(a)*\lim(b)$? so where does the error comes from i think it comes from the derivative step but I don't understand why

pie
  • 4,192
  • 1
    In the end, you'll need to prove the equivalent of the taylor series in some form. It will depend on how you define $e^x.$ – Thomas Andrews May 30 '23 at 17:28
  • 2
    You cannot take limits piecewise where some parts of the limit are evaluated but others are not - except in the cases where you can completely separate out the parts via factoring (multiplication) or addition from the entire limit. That clearly does not apply here. – Ninad Munshi May 30 '23 at 17:50
  • In the first equation, $\frac{2x}{x^3}$ is pulled out as $\frac1{x^2}$. – robjohn May 30 '23 at 18:13

6 Answers6

8

My comment addresses why your solution falls into a common problem solving pit trap. To do this without L'Hopital or series, consider that the limmand can be rewritten as

$$\frac{2\sinh x - 2x}{x^3} = 2 \cdot \frac{\sinh x - x}{x^3}$$

Why does this matter? Because hyperbolics follow similar identities to their trig counterparts. Consider the substitution $x = 3t$:

$$L = 2\lim_{x\to 0}\frac{\sinh x - x}{x^3} = 2 \lim_{t\to 0}\frac{\sinh 3t - 3t}{27t^3}$$

Using the identity $\sinh 3t = 3\sinh t + 4\sinh^3 t$, we obtain

$$L = \frac{2}{9}\lim_{t\to 0}\frac{\sinh t - t}{t^3} + \frac{8}{27}\lim_{t\to 0}\frac{\sinh^3 t}{t^3} = \frac{1}{9}L + \frac{8}{27}\left(\lim_{t\to 0}\frac{\sinh t}{t}\right)^3$$

$\frac{\sinh t}{t}$ is known limit, which we can prove via squeeze theorem equals $1$. This means we can now evaluate the final limit

$$L = \frac{1}{9}L + \frac{8}{27} \implies \boxed{L = \frac{1}{3}}$$

Ninad Munshi
  • 34,407
  • 3
    It is true only under the assumption that the limit $L$ exists and it is finite. The last statement is also valid, if $L=+\infty$ or $L=-\infty$. – Pavel R. May 30 '23 at 20:11
  • 1
    @PavelR. we are only computing a limit and not proving one. All methods are equivalent up to this point in getting the number. Proving the limit is this number (via standard epsilon delta) can only proceed once this "scratch work" is completed. In any case your comment would apply to any method we could use to solve this limit, be it series or L'Hopital. The purpose is to show that series and L'Hopital are not necessary and how to do it. – Ninad Munshi May 30 '23 at 20:24
  • 2
    I just want to show that the equation $L=L/9+8/27$ has more than one solution ($1/3$, $+\infty$, $-\infty$). Without any additional assumptions you cannot be sure that $L$ is just $1/3$. This ambiguity does not occur when you solve the limit using L'Hopital or a series expasion. – Pavel R. May 30 '23 at 20:53
4

Using the fine idea used here by robjohn, by binomial theorem we have

$$\left(1+\frac xn\right)^n-1-x-\frac{n-1}{2n}x^2 =\frac{(n-1)(n-2)}{6n^2}x^3+\sum_{k=4}^n\binom{n}{k}\frac{x^k}{n^k}$$

and for $|x|\le 1$

$$ \left|\sum_{k=4}^n\binom{n}{k}\frac{x^k}{n^k}\right| =|x|^4\left|\sum_{k=4}^n\binom{n}{k}\frac{x^{k-4}}{n^k}\right| \le |x|^4\sum_{k=4}^\infty\frac1{k!} =|x|^4\left(e-\tfrac83\right)$$

and combining this results we have that

$$\frac{e^x-1-x-\frac12 x^2}{x^3}=\frac16+O(x) \iff \frac{e^{-x}-1+x-\frac12 x^2}{-x^3}=\frac16+O(x)$$

and therefore

$$\frac{e^x-e^{-x}-2x}{x^3}=\frac{e^x-1-x-\frac12 x^2}{x^3}-\frac{e^{-x}-1+x-\frac12 x^2}{x^3} =\frac13+O(x) \to \frac13$$

user
  • 154,566
  • 1
    (+1) I had given a Binomial Theorem proof at the beginning of my answer, but decided to remove it so as not to confuse things. I had forgotten about the error estimate in the answer you quote. It does make the answer simpler. – robjohn Jun 02 '23 at 20:04
  • @robjohn Thanks! I think the way you have used there is a very nice and elegant way! I'd never thought about that without your fine example! Also the way you have proposed here is very nice., indeed(+1) already given. Bye – user Jun 02 '23 at 20:10
3

Integration by Parts $$ \begin{align} \lim_{x\to0}\frac{e^x-1}x &=\lim_{x\to0}\frac1x\int_0^xe^t\,\mathrm{d}t\tag{1a}\\ &=\lim_{x\to0}e^x-\lim_{x\to0}\frac1x \int_0^xte^t\,\mathrm{d}t\tag{1b}\\[3pt] &=1-0\tag{1c} \end{align} $$ Explanation:
$\text{(1a):}$ $e^x-1=\int_0^xe^t\,\mathrm{d}t$
$\text{(1b):}$ integration by parts
$\text{(1c):}$ $\left|\,\int_0^xte^t\,\mathrm{d}t\,\right|\le\frac12x^2e^{|x|}$

Next, integrate by parts three times: $$ \begin{align} &\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x^3}\tag{2a}\\[3pt] &=\lim_{x\to0}\frac1{x^3}\int_0^x\left(e^t+e^{-t}-2\right)\mathrm{d}t\tag{2b}\\ &=\lim_{x\to0}\frac{e^x+e^{-x}-2}{x^2}-\lim_{x\to0}\frac1{x^3}\int_0^xt\left(e^t-e^{-t}\right)\mathrm{d}t\tag{2c}\\ &=\lim_{x\to0}e^{-x}\left(\frac{e^x-1}x\right)^2-\lim_{x\to0}\left(\frac{e^x-e^{-x}}{2x}\right)+\lim_{x\to0}\frac1{x^3}\int_0^x\frac{t^2}2\left(e^t+e^{-t}\right)\mathrm{d}t\tag{2d}\\ &=\lim_{x\to0}e^{-x}\left(\frac{e^x-1}x\right)^2-\lim_{x\to0}e^{-x}\left(\frac{e^{2x}-1}{2x}\right)+\lim_{x\to0}\frac{e^x+e^{-x}}6\\ &-\lim_{x\to0}\frac1{x^3}\int_0^x\frac{t^4}3\left(\frac{e^t-e^{-t}}{2t}\right)\mathrm{d}t\tag{2e}\\[3pt] &=1-1+\frac13-0\tag{2f} \end{align} $$ Explanation:
$\text{(2b):}$ write $\text{(2a)}$ as an integral
$\text{(2c):}$ integrate by parts
$\text{(2d):}$ integrate by parts
$\text{(2e):}$ integrate by parts
$\text{(2f):}$ evaluate the limits using $(1)$ and
$\phantom{\text{(2f):}}$ $\left|\,\frac1{x^3}\int_0^x\frac{t^4}3\left(\frac{e^t-e^{-t}}{2t}\right)\,\mathrm{d}t\,\right|\le\frac{x^2}{15}\frac{\sinh(x)}x\le\frac{x^2}{15}\cosh(x)$

robjohn
  • 345,667
2

Substitute $x=\sqrt[3]{y}$ to get a derivative:

$$\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x^3} = 2 \lim_{y\to0} \frac{\sinh\left(\sqrt[3]{y}\right) - \sqrt[3]{y}}y = 2 \frac{d\left(\sinh\left(\sqrt[3]{y}\right) - \sqrt[3]{y}\right)}{dy}\bigg|_{y=0}$$

user170231
  • 19,334
  • 1
    You may continue after computing that derivative: $$ \frac{2}{3}\mathop {\lim }\limits_{y \to 0} \frac{{\cosh (\sqrt[3]{y}) - 1}}{{y^{2/3} }} = \frac{4}{3}\mathop {\lim }\limits_{y \to 0} \frac{{\sinh ^2 (\sqrt[3]{y}/2)}}{{y^{2/3} }} = \frac{1}{3}\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\sinh (\sqrt[3]{y}/2)}}{{\sqrt[3]{y}/2}}} \right)^2 = \frac{1}{3}. $$ – Gary Jun 01 '23 at 09:27
  • No, you may not: the question explicitly states that the solution should not use L'Hospital rule. So, this answer does not adress the question. – Anne Bauval Jul 04 '23 at 20:19
2

We have \begin{align*} \frac{{{\rm e}^x - {\rm e}^{ - x} - 2x}}{{x^3 }} & = 2\frac{{\sinh x - x}}{{x^3 }} = \frac{2}{{x^3 }}\int_0^x {(\cosh t - 1)\,{\rm d}t} \\ & = \frac{2}{{x^2 }}\int_0^1 {(\cosh (xt) - 1)\,{\rm d}t} = \frac{4}{{x^2 }}\int_0^1 {\sinh ^2 (xt/2)\,{\rm d}t} \\ & = \int_0^1 {t^2 \left( {\frac{{\sinh (xt/2)}}{{xt/2}}} \right)^2 {\rm d}t} \to \int_0^1 {t^2 \,{\rm d}t} = \frac{1}{3}, \end{align*} as $x\to 0$.

Gary
  • 31,845
0

$\lim (a+b)$ is not always equal to $\lim a + \lim b$. A necessary condition for that to be true is that $\lim a $ and $\lim b$ are finite. Same for multiplication.

Koobe
  • 376