If $$\lim\limits_{x\to 0} \left(2+\frac{f(x)}{x^2}\right)^{\frac 1x} = e^2$$ and $f(x)=a_0 + a_1x+a_2x^2...$, then prove that $$a_2+|a_3|=1\,.$$

The function seems to be of the form $\infty^{\infty}$.

I don’t know how to start solving it, since it an indeterminate form I don’t know how to evaluate. Can I gain some insight into the question?


$$y=\lim_{x\to 0} (2+\frac{f(x)}{x^2})^{\frac 1x}$$ $$\log y = \lim_{x\to 0} \frac 1x \log (2+\frac{f(x)}{x^2})$$ $$\log y = \frac 1x (1+\frac{f(x)}{x^2})=2$$

  • $\begingroup$ No it should be 1 raise to power infty $\endgroup$ Jul 26, 2020 at 12:01
  • 2
    $\begingroup$ What about looking at logarithms? $\endgroup$ Jul 26, 2020 at 12:07
  • 2
    $\begingroup$ You can take logarithms and make use of the standard limit $\lim\limits _{t\to 0}\dfrac{\log(1+t)}{t}=1$. $\endgroup$
    – Paramanand Singh
    Jul 26, 2020 at 13:13
  • $\begingroup$ From the limit condition, $$\left(2+\frac{f(x)}{x^2}\right)^{\frac1x}=\text{e}^2\big(1+o(1)\big)$$ for small values of $x$. Show that this means $$f(x)=-2x^2+x^2\,\text{e}^{2x}\,\big(1+o(x)\big)=-2x^2+x^2\,\big(1+2x+O(x^2)\big)\,\big(1+o(x)\big)$$ for small values of $x$. This implies $$f(x)=-x^2+2x^3+o(x^3)$$ for small values of $x$. Unless $f$ is analytic at $0$, I don't think you can expand $f$ further than this. $\endgroup$ Jul 26, 2020 at 14:10
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    $\begingroup$ @Batominovski: although this is not explicitly given, a proper interpretation could be that $f(x) $ satisfies the relation $f(x) =a_0+a_1x+a_2x^2+a_3x^3+o(x^3)$ as $x\to 0$. Based on this we can find the coefficients $a_0,a_1,a_2,a_3$ uniquely. $\endgroup$
    – Paramanand Singh
    Jul 26, 2020 at 14:42

2 Answers 2


This is not a difficult question, but a rigorous answer requires some effort and an understanding of limit laws.

As mentioned in my comment to the question I will assume that $$f(x) =a_0+a_1x+a_2x^2+a_3x^3+o(x^3)\tag{1}$$ as $x\to 0$. For those not well versed with the little-o notation the above equation is by definition equivalent to $$\lim_{x\to 0}\frac{f(x) - a_0-a_1x-a_2x^2-a_3x^3}{x^3}=0\tag{2}$$ The given limit condition in question is equivalent to $$\lim_{x\to 0}\frac{\log(2+(f(x)/x^2))}{x}=2\tag{3}$$ (this uses the fact that log is continuous and invertible).

And then $$\log\left(2+\frac{f(x)}{x^2}\right)=x\cdot\frac{\log(2+(f(x)/x^2))}{x}\to 0\cdot 2=0$$ as $x\to 0$. Exponentiating this we see that $$2+\frac{f(x)}{x^2}\to 1$$ as $x\to 0$ (this works because exponentiation is also continuous) or $$\frac{f(x)}{x^2}\to - 1\tag{4}$$ as $x\to 0$. Multiplying by $x$ and $x^2$ we see that $f(x) /x\to 0$ and $f(x) \to 0$.

Now using $(1)$ or $(2)$ with $f(x)\to 0$ we get $a_0=0$. Next using $a_0=0,f(x)/x\to 0$ with $(1)$ we get $a_1=0$. And in similar fashion we get $a_2=-1$ using $(1)$ and $(4)$.

To get $a_3$ we need to use equation $(3)$ also. Since the limit in $(3)$ is non-zero it follows from $(4)$ that $t=1+(f(x)/x^2)\to 0, t\neq 0$ as $x\to 0$ and $(3)$ can then be rewritten as $$\lim_{x\to 0}\frac{t}{x}\cdot\frac{\log(1+t)}{t}=2$$ Since $(\log(1+t))/t\to 1$ as $t\to 0$ it follows that $$\frac{t} {x} =\frac{1}{x}+\frac{f(x)}{x^3}\to 2\tag{5}$$ Using $(2)$ and $(5)$ along with the values $$a_0=a_1=0,a_2=-1$$ we get $a_3=2$.

The flaw in your approach is that you remove the limit operator without any justification. This simply does not work and can not be guaranteed.

  • $\begingroup$ Can you please clarify how you arrived at $2+\frac{f(x)}{x^2} \to 1$ $\endgroup$
    – Aditya
    Jul 27, 2020 at 7:07
  • $\begingroup$ @Aditya: I have proved that the log of this expression tends to $0$. Applying $\exp $ function on this we get that the expression tends to $1$. $\endgroup$
    – Paramanand Singh
    Jul 27, 2020 at 8:36
  • $\begingroup$ Why does the log tend to 0? $\endgroup$
    – Aditya
    Jul 27, 2020 at 9:33
  • $\begingroup$ @Aditya: read my answer after equation $(3)$. $\endgroup$
    – Paramanand Singh
    Jul 27, 2020 at 9:36
  • $\begingroup$ You multiplied and divided by $x$, and used $x\to 0$ to get the whole function $g(x)\to 0$. Doesnt that feel like cheating though? $\endgroup$
    – Aditya
    Jul 27, 2020 at 12:03

Lemma: If $\lim_\limits{x\rightarrow 0}(1+g(x))^{\frac{1}{x}}=e^2$, then $\lim_\limits{x\rightarrow 0}g(x)=0$.

Proof of the Lemma: First, we know that $1+g(x)>0$ (see:https://www.wolframalpha.com/input/?i=%281%2Bx%29%5E%7B1%2Fx%7D)

If $\lim_\limits{x\rightarrow 0}g(x)\neq 0$, by definition of limit, there exists $\epsilon_0>0$ and sequence $x_n\rightarrow 0$ and a small enough $\delta$ (say $\delta=10^{-10}$, anyway) when $|x_n|<\delta$, we have $g(x_n)\geq \epsilon_0$ or $g(x_n)\leq -\epsilon_0$.

Now we consider the part of $\delta>x_n>0$ to construct a contradiction.

(i) if $g(x_n)\geq \epsilon_0$, then $$(1+g(x_n))^{\frac{1}{x_n}}\geq(1+\epsilon_0)^{\frac{1}{x_n}}\geq(1+\epsilon_0)^{\delta^{-1}}>e^2$$ ($t^{1/x_n}$ is increasing, $\delta$ small enough)

(ii) if $g(x_n)\leq -\epsilon_0$ $$0<(1+g(x_n))^{\frac{1}{x_n}}\leq(1-\epsilon_0)^{\frac{1}{x_n}}\leq(1-\epsilon_0)^{\delta^{-1}}<e^2$$ ($t^{1/x_n}$ is increasing, $\delta$ small enough)

Contradiction. The lemma is proved.

Now back to the problem. By the lemma we have $\frac{f(x)}{x^2}+1$ goes to $0$, that is $$\frac{a_0}{x^2}+\frac{a_1}{x}+a_2+a_3x+o(x)+1\rightarrow 0$$ So $a_0=a_1=0$, $a_2=-1$. Now $$(2+\frac{f(x)}{x^2})^{1/x}=(1+a_3x+o(x))^{1/x}=\\ (1+a_3 x+o(x))^{\frac{1}{a_3 x+o(x)}\cdot~~~~~\frac{a_3 x+o(x)}{x}}\rightarrow e^{a_3}=e^2$$

So $a_3=2$


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