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

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?

EDIT:-

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

• No it should be 1 raise to power infty Jul 26, 2020 at 12:01
• What about looking at logarithms? Jul 26, 2020 at 12:07
• You can take logarithms and make use of the standard limit $\lim\limits _{t\to 0}\dfrac{\log(1+t)}{t}=1$. Jul 26, 2020 at 13:13
• 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. Jul 26, 2020 at 14:10
• @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. Jul 26, 2020 at 14:42

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.

• Can you please clarify how you arrived at $2+\frac{f(x)}{x^2} \to 1$ Jul 27, 2020 at 7:07
• @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 $1$. Jul 27, 2020 at 8:36
• Why does the log tend to 0? Jul 27, 2020 at 9:33
• @Aditya: read my answer after equation $(3)$. Jul 27, 2020 at 9:36
• 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? 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 $$g(x_n)\leq -\epsilon_0$$ $$0<(1+g(x_n))^{\frac{1}{x_n}}\leq(1-\epsilon_0)^{\frac{1}{x_n}}\leq(1-\epsilon_0)^{\delta^{-1}} ($$t^{1/x_n}$$ is increasing, $$\delta$$ small enough)

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 $$a_3=2$$