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I am stuck on how to calculate this limit:

$$\displaystyle{\lim_{x \to \infty}}{x\big(e^3 - \big(1 + \frac{1}{x}\big)^{3x}\big)}$$

Honestly, I don't know where to even start there, or how to simplify this expression.

What I see limit can be rewritten also to this form:

$$\displaystyle{\lim_{x \to \infty}}{x\big(e^3 - e^{3x\log{\big(1 + \frac{1}{x}\big)}}\big)}$$

but don't know how to continue with that.

Thanks.

meerkat
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4 Answers4

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Let $$g(h)=(1+h)^{3/h}$$ when $h>0$ and $e^3$ when $h=0.$

Then $g(h)$ is continuous on $[0,+\infty)$ and differentiable on $(0,+\infty).$

If $x=1/h$ for $h>0$ then:

$$x\left (e^3-\left(1+\frac1x\right)^{3x}\right)=\frac{g(0)-g(h)}{h}=-g'(c)$$ for some $c\in(0,h).$

Now, $$g'(c)=3(1+c)^{3/c}\cdot\frac{\frac{c}{c+1}-\log(1+c)}{c^2}$$

You can compute the limit of the fractional part, $L,$ as $c\to0$ using L'Hopital, and you get your limit is $-3e^3L.$

A little easier if you know the power series:

$$\frac{c}{c+1}=c-c^2+o(c^2)\\ \ln(1+c)=c-\frac{c^2}2+o(c^2)$$ From this, we get $L=-\frac12.$

Thomas Andrews
  • 177,126
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If you set $z=\frac1x$ $$\log(1+z) = z - \frac{z^2}2 + \frac{z^3}3 - o(z^4)$$ Then applying l'Hopital $$\lim_{z\to0}\frac{e^3 - e^{3\frac1{z}(z - \frac{z^2}2 + \frac{z^3}3 - o(z^4))}}{z} = $$ $$ = \lim_{z\to0}\frac{e^3 - e^{3(1 - \frac{z}2 + \frac{z^2}3 - o(z^3))}}{z} = $$ $$ = -\lim_{z\to0}e^{3(1 - \frac{z}2 + \frac{z^2}3 - o(z^3))}\cdot3(-\frac12 + \frac{2z}3-\frac{d}{dz}o(z^3)) = \frac32e^3$$

Sergey Zaitsev
  • 765
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Hint: Put $\dfrac{1}{x} = z$

The limit will transpose to $$\lim_\limits{z \to 0} \dfrac{e^3 - (1+z)^{3/z}}{z} = \lim_\limits{z \to 0} \dfrac{e^3 - e^{3\log(1+z)^{1/z}}}{z}$$

SchrodingersCat
  • 24,472
  • Yeah, I've tried that and also applied l'Hopital once, but problem is, e function will stay in numerator so I am not sure what to do next after that. Also small detail as x goes to inf - I think this should translate to z -> 0+. – meerkat Jan 29 '23 at 20:10
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Here's my answer:

$$\begin{aligned} \displaystyle{ \lim_{x \to \infty}}{x\big(e^3 - e^{3x\log{\big(1 + \frac{1}{x}\big)}}\big)} &= \lim_{h \to 0+}{\frac{e^3 - e^{\frac{3\log{(1 + h)}}{h}}}{h}} \\ &= -e^3\lim_{h \to 0+}{\frac{e^{\frac{3\log{(1 + h)}}{h}} - 1}{\frac{3\log{(1 + h)}}{h}}\frac{\frac{3\log{(1 + h)}}{h}}{h}} \\ &= -e^3\lim_{h \to 0+}{\frac{\frac{3}{1 + h} - 3}{2h}} \\ &= -e^3*\frac{-3}{2} = \frac{3e^3}{2} \end{aligned}$$

This is most simple in my opinion.

First, we substitute $h = \frac{1}{x}$, then $\displaystyle{\lim_{x \to 0+}{\frac{3log(1 + x) - 3x}{x}} = 0}$ (second line).

And $\displaystyle{\lim_{x \to 0+}{\frac{e^{x} - 1}{x}} = 1}$ (second line).

Applying L'Hôpital's rule (second to third line).

Finally, using arithmetic limit, we get $\frac{3e^3}{2}$.

Credit: Martin R for mentioning this problem with solution where I got an inspiration.

meerkat
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