Find the two limits without the use of l'Hospital's rule or series expansion.

Question

I was asked to evaluate these two limits:

$$\lim_{x\rightarrow0}\frac{x^3}{x-\sin x}$$ $$\lim_{x\rightarrow0}\frac{e^{-x^2}+x^2-1}{\sin(3x^4)}$$

For the first one I tried to divide the numerator and denominator by $x^3$, but I can't get the answer unless I apply l'Hospital's rule or using a series expansion.

I also tried to use a substitution $u=x^2$ for the second limit, but I can't seem to relate anything between the exponential function and sine function.

Answer 1

As shown here we have that

• $\lim_{x\to0}\frac{\sin x-x}{x^3}=-\frac16$
• $\lim_{x\to0}\frac{e^x-x-1}{x^2}=\frac12$

For the second one we can use that

$$\frac{e^{-x^2}+x^2-1}{\sin(3x^4)}=\frac{e^{-x^2}+x^2-1}{-x^4}\cdot \frac{-3x^4}{\sin(3x^4)}\cdot\frac13$$

Answer 2

Write $$0<\cos x<1$$ for $0<x<\dfrac\pi2$ and integrate five times from $0$ to $x$. This will establish $$0<\sin x-x+\frac{x^3}6<\frac{x^5}{120}$$ and the first limit easily follows.

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The plot shows

$$x-\frac{x^3}6<\sin x<x-\frac{x^3}6+\frac{x^5}{120}$$

Of course this matches the terms of Taylor's expansion, but without explicitly using the theorem.

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For the second limit, you can reduce to

$$\lim_{t\to0}\frac{e^t-t-1}{3t^2}.$$

And for $t\in(0,1)$,

$$1<e^t<e$$

yields afer three integrations from $0$ to $t$ $$1+t+\frac{t^2}2+\frac{t^3}6<e^t<1+t+\frac{t^2}2+e\frac{t^3}6$$

Answer 3

Introduction (can skip). This answer aims at producing a solution which rely on as less knowledge as possible. This does not necessarily mean that this solution is easy. Indeed, if we allow more advanced tools, then much shorter proofs are available. Here are some examples:

1. Integrating $-1 \leq \sin(x) \leq 1$ four times yields $-\frac{x^4}{4!} \leq \sin(x) - \big( x-\frac{x^3}{3!} \big) \leq \frac{x^4}{4!}$. This immediately yields the limit via squeezing.

2. Using $\limsup$ and $\liminf$, we can bypass the question of existence of limit. This allows to directly adopt the arguments in this answer.

3. Not to mention, L'Hospital's rule or Taylor's theorem gives a one-liner proof.

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Solution. In this answer, we will only use following properties of trigonometric functions.

• $\text{(P1)} \ $ $0 \leq \sin x \leq x \leq \tan x$ for any $0 < x < \frac{\pi}{2}$.

• $\text{(P2)} \ $ $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(2x) = 1 - 2\sin^2(x)$.

• $\text{(P3)} \ $ $0 < \cos(y) \leq \cos(x) \leq 1$ for $0 < x < y < \frac{\pi}{2}$.

• $\text{(P4)} \ $ $\sin(-x) = -\sin(x)$.

Define $f(x)$ by

$$f(x) = \frac{x - \sin(x)}{x^3}$$

for $x \neq 0$. From $\text{(P4)}$, we have $f(-x) = f(x)$ and hence it suffices to examine the right-limit of $f$. We also write $\operatorname{sinc}(x) = \sin(x)/x$ for simplicity. By the double-angle formula $\text{(P2)}$,

\begin{align*} f(x) &= \frac{x - 2\sin(x/2)}{x^3} + \frac{2\sin(x/2)\left(1 - \cos(x/2) \right)}{x^3} \\ &\quad= \frac{1}{4}f\left(\frac{x}{2}\right) + \frac{\operatorname{sinc}(x/2)\operatorname{sinc}^2(x/4)}{8} \\ &\quad\quad \ldots \\ &\quad\quad\quad = \frac{1}{4^n}f\left(\frac{x}{2^n}\right) + \frac{1}{2} \sum_{k=1}^{n} \frac{\operatorname{sinc}(x/2^k)\operatorname{sinc}^2(x/2^{k+1})}{4^k}. \end{align*}

Now assume that $x \in (0, \frac{\pi}{2})$. Then by $\text{(P1)}$ and $\text{(P2)}$,

$$ 0 \leq f(x) \leq \frac{\tan(x)-\sin(x)}{x^3} = \frac{\sin(x) (1 - \cos(x))}{x\cos(x)} = \frac{2\sin(x)\sin^2(\frac{x}{2})}{x^3\cos(x)} \leq \frac{1}{2\cos(x)}. \tag{2} $$

Also, by $\text{(P1)}$ and $\text{(P3)}$,

$$ 1 \geq \operatorname{sinc}(x/2^n) = \frac{\tan(x/2^n)}{x/2^n}\cos(x/2^n) \geq \cos(x). \tag{3} $$

Using $\text{(2)}$, $\text{(P3)}$, and $\text{(3)}$, we can bound each term in $\text{(1)}$ from above and below, for $x \in (0, \frac{\pi}{2})$. Indeed, we get

$$ \frac{1}{2\cos^3(x)} \sum_{k=1}^{n} \frac{1}{4^k} \leq f(x) \leq \frac{1}{4^n} \frac{1}{2\cos(x)} + \frac{1}{2} \sum_{k=1}^{n} \frac{1}{4^k}. $$

Since $f(x)$ is independent of $n$, letting $n\to\infty$ and using the geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$ for $|r| < 1$ shows that

$$ \frac{1}{6\cos^3(x)} \leq f(x) \leq \frac{1}{6}. $$

By the squeezing theorem, $f(x) \to \frac{1}{6}$ as $x \to 0^+$. (Here, we used the fact that $\cos(x) \to 1$ as $x \to 0$, which itself can be proved from $\cos(x) = 1-2\sin^2(x/2) \in [1-\frac{x^2}{2}, 1]$ together with the squeezing theorem.)