0

The limit of this function is 2, seen from graphing calculators. My friend said he could only solve it using L'Hopital's rule or taylor expansion, and me and him were wondering if it was possible to solve it without these 2 methods (these methods are not taught at school). Interestingly, I found different ways of calculating it, and got 0 and 1, but not 2. Here are my ways of solving it:

$$\lim_{x \to 0} \frac{e^{2x}-1}{x^{2}}-\frac{2x}{x^{2}}$$ $$=2\lim_{x \to 0} \frac{e^{2x}-1}{2xx}-\frac{2x}{2xx}$$ $$=2\lim_{x \to 0} \frac{1}{x}-\frac{1}{x}$$ $$=0$$ Or, $$=\lim_{x \to 0} \frac{(e^{x}-1)(e^{x}+1)}{x^{2}}-\frac{2x}{x^{2}}$$The above part still equals to 2 using a calculator. However, I cannot find the problem as the bottom parts equal to 1 and I'm not sure why. $$=\lim_{x \to 0} \frac{e^{x}+1}{x}-\frac{2}{x}$$ $$=1$$ Any help in pointing out my errors are welcome.

hablahblahha
  • 89
  • 4
    Replacing the limit in one part but not the other usually doesn't lead to correct results. Review your courses on when $ \lim\left(f+g\right)=\lim f + \lim g $, and $ \lim\left(fg\right)=\lim f \times\lim g $. – CHAMSI Jun 04 '25 at 12:02
  • 1
    Generally, I disapprove of such stunts: you do professional maths with professional tools. There are some excuses, mainly contests where the contestants don't know those professional tools, yet. In this case, you should explain what can be used. You can't even define $e$ with grade school maths, let alone $e^x$, or any limit. So what is the toolset you'll gracefully allow? – wasn't me Jun 04 '25 at 12:30
  • Your evaluation attempts of the limits are wrong. They are indeterminate forms. Using L' Hopital's rule yields the right limit. – Kirusanth Jun 04 '25 at 14:39
  • Refer https://en.m.wikipedia.org/wiki/Indeterminate_form – Kirusanth Jun 04 '25 at 14:48

3 Answers3

2

Well, if you do not define the exponential function as its Taylor series, I assume you define it as below. In Real Analysis, it is often standard or convenient to define $\exp(x)$ as its Taylor Series. It would be helpful to know whether you're in a Calculus class or doing Real Analysis.

We want to compute the limit:

$$ \lim_{x \to 0} \frac{e^{2x} - 2x - 1}{x^2} $$

We will use the limit definition of the exponential function:

$$ e^y := \lim_{n \to \infty} \left(1 + \frac{y}{n} \right)^n $$

Applying this to $y = 2x$, define:

$$ f_n(x) = \left(1 + \frac{2x}{n} \right)^n \quad \text{so that} \quad e^{2x} = \lim_{n \to \infty} f_n(x) $$

Our goal is to compute:

$$ \lim_{x \to 0} \frac{\lim_{n \to \infty} f_n(x) - 2x - 1}{x^2} $$

Using the binomial theorem:

$$ f_n(x) = \sum_{k=0}^n \binom{n}{k} \left( \frac{2x}{n} \right)^k $$

Let's look at the first few terms:

  • $k = 0$: $ \binom{n}{0} \left(\frac{2x}{n}\right)^0 = 1 $
  • $ k = 1 $: $ \binom{n}{1} \left(\frac{2x}{n}\right) = 2x$
  • $ k = 2 $:
    $$ \binom{n}{2} \left(\frac{2x}{n}\right)^2 = \frac{n(n-1)}{2} \cdot \frac{4x^2}{n^2} = 2x^2 \left(1 - \frac{1}{n} \right) $$

So we write:

$$ f_n(x) = 1 + 2x + 2x^2 \left(1 - \frac{1}{n} \right) + R_n(x) $$

where

$$ R_n(x) := \sum_{k=3}^n \binom{n}{k} \left( \frac{2x}{n} \right)^k $$

Now take the limit as $n \to \infty$:

$$ \lim_{n \to \infty} f_n(x) = 1 + 2x + 2x^2 + \lim_{n \to \infty} R_n(x) $$

We now estimate $R_n(x)$. Since:

$$ |R_n(x)| \le \sum_{k=3}^n \binom{n}{k} \left( \frac{2|x|}{n} \right)^k $$

and for fixed small $x$, all terms in this sum are $O(x^3)$, it follows that:

$$ R_n(x) = o(x^2) $$

Hence:

$$ e^{2x} = 1 + 2x + 2x^2 + o(x^2) $$

Then:

$$ e^{2x} - 2x - 1 = 2x^2 + o(x^2) $$

and

$$ \frac{e^{2x} - 2x - 1}{x^2} = 2 + \frac{o(x^2)}{x^2} $$

Taking the limit as $x \to 0$:

$$ \boxed{\lim_{x \to 0} \frac{e^{2x} - 2x - 1}{x^2} = 2} $$

If you're not familiar with the Big-O-Notation, check out Wikipedia for a good explanation. Alternatively, further expand the binomial and omit the Big-O-Notation...

Samuel
  • 1,051
  • So you are using an expansion against the OP's request. $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \cdots=1 + 2x + o(x^2)$. – Sebastiano Jun 04 '25 at 12:40
  • 2
    I use the binomial expansion. OP requested no Taylor expansion. Nothing wrong with that. – Samuel Jun 04 '25 at 12:42
  • I did not write that your answer is wrong. – Sebastiano Jun 04 '25 at 12:42
  • Indeed, but you suggest I didn't properly read OP's question. Surely, those series turn out to be equal, but this solution uses Taylor not even once. – Samuel Jun 04 '25 at 12:45
  • I know that: Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be a function, $x_0 \in ]a, b[$, and suppose that the function is differentiable $(n-1)$ times in a neighborhood $I$ of $x_0$. If the $n$-th derivative exists at the point $x_0$, i.e., $f^{(n)}(x_0)$, the Taylor series with Peano remainder of order $n$ holds: $ f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k + o\left( (x - x_0)^n \right) $. The last term in the equality is called the Peano remainder of order $n$: $ R_n(x) = o\left( (x - x_0)^n \right). $ – Sebastiano Jun 04 '25 at 12:51
  • 2
    I agree with Sebastiano, that when you are using Taylor polynomial, then it's mainly same with Taylor expansion. – zkutch Jun 04 '25 at 12:51
  • I took a shortcut around using that theorem via the binomial theorem. Obviously, any expansion will be equivalent. There are no Taylor polynomials directly involved. – Samuel Jun 04 '25 at 12:55
0

I am only assuming one knows $\frac{\mathrm d}{\mathrm dx}\operatorname e^x=\operatorname e^x$ (so $\operatorname e^x$ is an antiderivative of itself) and $\operatorname e^0=1$. Then you can start with $$\left\lvert\operatorname e^x-1\right\rvert\le\varepsilon$$ for $x$ sufficiently close to $0$ and proceed with repeated integration (and triangle inequality for the integral): \begin{gather*} \left\lvert\operatorname e^x-1-x\right\rvert=\left\lvert\int_0^x(\operatorname e^t-1)\,\mathrm d t\right\rvert\le\varepsilon |x|,\\[.4em] \left\lvert\operatorname e^x-1-x-\frac{x^2}2\right\rvert\le\varepsilon \frac{x^2}2, \end{gather*} when $x$ is sufficiently close to $0$. Substituting $2x$ to $x$ gives: $$\left\lvert\operatorname e^{2x}-1-2x-2x^2\right\rvert\le2\varepsilon x^2.$$ Divide by $x^2$ to get $$\limsup_{x\to0}\left\lvert\frac{\operatorname e^{2x}-1-2x}{x^2}-2\right\rvert\le2\varepsilon,$$ and let finally $\varepsilon\to0^+$.

nejimban
  • 4,107
0

Let $ x\in\mathbb{R} $. One can prove using IBP that : \begin{aligned} \frac{\mathrm{e}^{2x}-2x-1}{x^{2}}&=4\int_{0}^{1}{\left(1-y\right)\mathrm{e}^{2xy}\,\mathrm{d}y}\\&=2+4\int_{0}^{1}{\left(1-y\right)\left(\mathrm{e}^{2xy}-1\right)\mathrm{d}y} \end{aligned}

Using the inequality $ \left\lvert\mathrm{e}^{x}-1\right\rvert\leq\left\lvert x\right\rvert $, $$ \left\lvert\int_{0}^{1}{\left(1-y\right)\left(\mathrm{e}^{2xy}-1\right)\mathrm{d}y}\right\rvert\leq\int_{0}^{1}{\left(1-y\right)\left\lvert\mathrm{e}^{2xy}-1\right\rvert\mathrm{d}y}\leq 2\left\lvert x\right\rvert\int_{0}^{1}{y\left(1-y\right)\mathrm{d}y}=\frac{\left\lvert x\right\rvert}{3}\underset{x\to 0}{\longrightarrow}0 $$

Thus, the limit is $ 2 $.

CHAMSI
  • 9,222