# How to compute $\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2}$ without L'Hospital's Rule or series?

I came across this problem in an examination for beginners in Calculus:

Compute $\displaystyle \lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2}$.

It is easy to show that the limit is $1$ by using series or L'Hospital's Rule. But the examination is aimed to students that know only the basics of Calculus (derivatives of power functions and trigonometric functions, Chain Rule, Mean Value Theorem basically).

How to compute this limit by elementary means?

• $\sin(x^2+x)=x+x^2+\mathcal{O}(x^3)$ works, but that is also excluded I assume? – Fritz Oct 15 '17 at 8:19
• @MarvinF: unfortunately, yes. – Taladris Oct 15 '17 at 8:20
• can you use $sin(x) \sim x?$ – MiKiDe Oct 15 '17 at 8:37
• Good question +1. And use of advanced tools is not necessary here. – Paramanand Singh Oct 15 '17 at 11:54

\begin{align} L = &\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2} &\\ = &\lim_{x\to 0}\frac{\sin x^2 \cos x + \sin x \cos x^2-x}{x^2} &\\= &1+ \lim_{x\to 0}\frac{ \sin x \cos x^2-x}{x^2}&\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x + x) \cos x^2-x}{x^2}&\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x)\cos x^2 + x(\cos x^2 -1)}{x^2} &\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x)\cos x^2 - x(2\sin^2(x^2/2))}{x^2} \tag{0} &\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x)}{x^2} \end{align}

Now we need to find $\lim_{x\to 0}\dfrac{ (\sin x - x)}{x^2}$

We know that $\lim_{x\to 0} \dfrac{\sin x - x}{x^3}$ is $\dfrac {-1}{6}$,

Therefore,

$$\lim_{x\to 0}\frac{ (\sin x - x)}{x^2} = \lim_{x \to 0} x \cdot \lim_{x\to 0}\frac{ (\sin x - x)}{x^3} = 0 \cdot \dfrac{-1}{6} = 0$$

Edit :-

An alternative method to prove $\lim_{x\to 0}\dfrac{ (\sin x - x)}{x^2} = 0$ is given by Paramanand Singh in the comments,

$$\cos x \leq \dfrac{\sin x}{x} \leq 1 \implies \dfrac{\cos x - 1}{x}\leq \dfrac{\sin x - x}{x^2} \leq 0$$

Using seqeeze theorem and the fact that $\lim_{x\to 0}\dfrac{\cos x - 1}{x} = 0$, we get $\lim_{x\to 0}\dfrac{ (\sin x - x)}{x^2} = 0$.

Therefore $L =1$

$1) :-$ I used the following on step $(0)$, $$\lim_{x\to 0} \dfrac{x\sin^2 (x^2/2)}{x^2} =\lim_{x\to 0} x\left(\dfrac{\sin (x^2/2)}{x}\right)^2 =\lim_{x\to 0} x\left(\dfrac{\sin (x^2/2)}{x^2} x\right)^2 = 0$$

$2):-$ $\lim_{x\to 0}\dfrac{\sin x - x}{x^3}$ is derived here. (This limit is not required anymore to find $L$)

• You may directly prove that $(\sin x-x) /x^{2}\to 0$ using Squeeze theorem and inequality $\cos x<(\sin x) /x<1$. Apart from that your solution is simplest (looks longer because of detailed steps). +1 – Paramanand Singh Oct 15 '17 at 11:36
• @ParamanandSingh Thank you very much for the upvote. Your method is certainly far superior to that of mine. – user8277998 Oct 15 '17 at 12:05
• If you show that$$\lim_{x\to0}\frac{\cos(x)-1}x$$exists (by noting the derivative of $\cos$ exists, for example), then it may be computed trivially by letting $x=-t$ and doing some limit algebra. – Simply Beautiful Art Oct 15 '17 at 12:25
• @SimplyBeautifulArt What is the need of doing it ? It can be proved using $\cos x - 1 = -2 \sin^2(x/2)$. – user8277998 Oct 15 '17 at 12:31
• :o That certainly works too. :D – Simply Beautiful Art Oct 15 '17 at 12:35

First we prove that $\lim_{x\to 0} \frac{\sin x - x}{x^2} =0$, which can be done similarly as in the case of proving that $\lim_{x\to 0} \frac{\sin x}{x} =1$. We have that:

$$cos(x) \le \frac{\sin x}{x} \le 1 \implies \frac{\cos x - 1}{x} \le \frac{\sin x - x}{x^2} \le 0$$

and so by the Squeeze Theorem the claim is proven, using the well-known $\lim_{x\to 0} \frac{\cos x - 1}{x} =0$

Now using the MVT we have:

$$\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2} = \lim_{x \to 0} \frac{\sin x - x}{x^2} +\lim_{x \to 0} \frac{\sin(x^2+x) - \sin(x)}{(x^2 + x) - x}= 0 + \lim_{x \to 0} \cos(c)= \cos(0) = 1$$

As $c \in (x,x^2+x)$ and $x \to 0$

• The replacement of $x$ by $\sin x$ is invalid and it amounts to showing that $(\sin x-x) /x^{2}\to 0$ as $x\to 0$. Such practice should be avoided. - 1 – Paramanand Singh Oct 15 '17 at 11:40
• @ParamanandSingh I appreciate following the downvote with an explanation. You're right about that, as I implicitly use that $\sin(x) = x + o(x^3)$, which I doubt that the OP can use. Anyway it's not hard to prove the needed limit. I will do it now and fix the answer – Stefan4024 Oct 15 '17 at 11:59
• Downvote converted into an upvote. +1 Glad that you took my feedback in positive manner. There is always a risk of high tempers when explaining a downvote. But I prefer to take that risk. – Paramanand Singh Oct 15 '17 at 12:56
• @ParamanandSingh I appreciate your comment. In fact I believe that if you downvote you should give a proper explanation why you did it, pointing out mistakes in the proof etc. I would welcome a constructive negative critic than a dull downvote without any reply. After all we all came here to learn and improve ourselves. Anyway, thanks for pointing out the mistake and keep up the good work. – Stefan4024 Oct 15 '17 at 13:02
• A non-mathematical theorem: every significant contributor on this website has been a victim of unexplained downvotes at least once (you and myself included). – Paramanand Singh Oct 15 '17 at 13:06

Let $u=x^2+x$ and see that

\begin{align}\frac{\sin(x^2+x)-x}{x^2}&=\frac{\sin(x^2+x)-(x^2+x)+x^2}{(x^2+x)^2}(x+1)^2\\&=\frac{\sin(u)-u}{u^2}(x+1)^2+1\\&\to0+1=1\end{align}

where we used

$$\lim_{u\to0}\frac{\sin(u)-u}{u^2}=0$$

(various proofs see here)

• The derivative $f'(x)$ for $x\neq 0$ can be evaluated using derivative rules. Evaluation of $f'(0)$ is not possible in thisanner and it essentially requires you to evaluate the limit in question. You may perhaps use MVT to write the difference quotient and a derivative $f'(\xi)$ for some $\xi$ between $0$ and $x$ and then take limit of $f'(x)$ as $x\to 0$ and this part is complicated. – Paramanand Singh Oct 15 '17 at 11:43
• Oh pooey, you are right. I'd need to show $f$ is continuously differentiable for this to work... – Simply Beautiful Art Oct 15 '17 at 11:46
• Don't worry, it happens. But I expect you to come up with another solution. – Paramanand Singh Oct 15 '17 at 11:47
• :P Of course you do. Well, I'll be thinking on it. – Simply Beautiful Art Oct 15 '17 at 11:48
• How about now? @ParamanandSingh – Simply Beautiful Art Oct 15 '17 at 12:17