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Is it possible to solve this without using L'Hôpital's rule?

$$ \lim_{x\to 0} \Big(\frac{3x+1}{x}-\frac{1}{\sin x}\Big) $$

I tried to solve it but I got stuck at the $\frac{1}{\sin x}$ part.

$$ =\lim_{x\to 0}\frac{3+x^{-1}}{1}-\lim_{x\to 0}\frac{1}{\sin x} $$

since $\lim_{x\to 0}\frac{1}{\sin x}$ does not exist.

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2  
$\lim\limits_{x\to 0}x^{-1}$ doesn't exist either. –  Jonas Meyer Jan 29 '13 at 3:03
    
You cannot take the limit of the two terms separately. Instead combine them into one fraction. At some point you will need to find $\lim_{x\to0}\frac{\sin x}{x}$ and don't think there is any way to do this without L'Hopital's rule (or the definition of derivative of $\sin$ at the origin) without rederiving the limit from first principles --- see any Calculus textbook. –  kiwi Jan 29 '13 at 3:04
    
For what it's worth: We have $\dfrac1{x}-\dfrac1{\sin x}=\dfrac{\dfrac{\sin x}{x}-1}{x}\cdot\dfrac{x}{\sin x}=\dfrac{f(x)-f(0)}{x}\cdot\dfrac{1}{f(x)},$ where $f$ is the sinc function $f(x)=\dfrac{\sin x}{x}$, $x\neq 0$, $f(0)=1$. Since $\lim\limits_{x\to 0}f(x)=f(0)=1$ and $f'(0)=0$, and the limit of a product is the product of the limits, the limit of the expression above is $0$. (But this only reduces the problem to showing that $f'(0)$ exists. If we know that $f'(0)$ exists it follows from oddness that $f'(0)=0$.) –  Jonas Meyer Jan 29 '13 at 3:32
    
@JonasMeyer - This is getting more complicated than I expected... –  Derek 朕會功夫 Jan 29 '13 at 3:34

6 Answers 6

up vote 12 down vote accepted

Via power series

$$\frac{3x+1}{x}-\frac{1}{\sin x}=\frac{3x\sin x+\sin x-x}{x\sin x}=$$

$$\frac{3x\left(x-O(x^3)\right)+\left(x-O(x^3)\right)-x}{x\left(x-O(x^3)\right)}=\frac{3x^2-O(x^4)-O(x^3)}{x^2-O(x^4)}=$$

$$3\frac{1-O(x^2)-O(x)}{1-O(x^2)}\xrightarrow [x\to 0]{}3\cdot 1=3$$

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When $x\ne 0$, our function is equal to $$3-\left(\frac{x-\sin x}{x\sin x}\right).$$ So we want to find $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}.$$ This limit is obviously $0$ (look at the Maclaurin series of $\sin x$). But presumably if L'Hospital's Rule is forbidden, then Maclaurin series are also forbidden.

We consider positive $x$ only, since if we deal with these, what happens for negative $x$ is clear by symmetry.

We need to know something about the behaviour of $x-\sin x$ for positive $x$ near $0$.

$1$. First we show that $x\ge \sin x$ if $x\ge 0$. Let $f(x)=x-\sin x$. Note that $f(0)=0$ and $f'(x)=1-\cos x\ge 0$, so $f(x)$ is non-decreasing for positive $x$. It follows that $f(x)\ge 0$ if $x\ge 0$.

$2$. Next we show that $\cos x\ge 1-\frac{x^2}{2}$ for $x\ge 0$. Let $g(x)=\cos x-\left(1-\frac{x^2}{2}\right)$. Note that $g(0)=0$ and, by the result of $(1)$, $g'(x)=-\sin x+x\ge 0$ for $x\ge 0$. It follows that $g(x)\ge 0$ for $x\ge 0$.

$3$. Next we show that $\sin x\ge x-\frac{x^3}{6}$ for $x\ge 0$. Let $h(x)=\sin x-\left(x-\frac{x^3}{6}\right)$. We have $h(0)=0$, and by the result of $(2)$, $h'(x)\ge 0$ if $x\ge 0$. The desired result follows.

We conclude thaat if $x\ge 0$, then $0\le x-\sin x\le \frac{x^3}{6}$. From this inequality, it follows easily that $\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=0$.

Remark: Any proof must address the behaviour of $x-\sin x$ near $0$. Of course we do not need the very strong $|x-\sin x|\le \frac{x^3}{6}$ that we obtained above. But we must show at least that $|x-\sin x|=o(x^2)$.

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Here is a hint which uses no L'Hôpital, no derivatives, and no power series; just trigonometry.

Hint: In this answer, it is shown, geometrically, that $\sin(x)\le x\le\tan(x)$ for $x\in[0,\pi/2)$. Thus, $$ \begin{align} \left|\,\frac{1}{x}-\frac{1}{\sin(x)}\,\right| &=\left|\,\frac{\sin(x)-x}{x\sin(x)}\,\right|\\ &\le\left|\,\frac{\sin(x)-\tan(x)}{x\sin(x)}\,\right|\\ &=\left|\,\frac{1-\sec(x)}{x}\,\right|\\ &=\left|\,\frac{1-\sec(x)}{x}\frac{1+\sec(x)}{1+\sec(x)}\,\right|\\ &=\left|\,\frac{1-\sec^2(x)}{x(1+\sec(x))}\,\right|\\ &=\left|\,\frac{-\tan^2(x)}{x(1+\sec(x))}\,\right|\\ &=\left|\,\frac{\tan(x)}{x}\frac{\tan(x)}{1+\sec(x)}\,\right|\\ &=\left|\,\frac{\tan(x)}{x}\tan(x/2)\,\right| \end{align} $$

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Let $\displaystyle L=\lim_{x \to 0} \frac{\sin x - x}{x^2}$. Replacing $x$ by $2y$, we obtain:

$ \begin{align*} L & = \lim_{y \to 0} \frac{\sin 2y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2 \sin y + 2 \sin y - 2y}{4y^2} \\ & = \lim_{y \to 0} \left( \frac{2 \sin y(\cos y-1)}{4y^2}+\frac{1}{2} \lim_{y \to 0} \frac{\sin y - y}{y^2}\right) \\ & = \frac{1}{2} \lim_{y \to 0} \sin y \cfrac{-2\sin^2\cfrac{y}{2}}{y^2}+\frac{1}{2}L \\ & = -\frac{1}{4} \lim_{y \to 0} \sin y \cfrac{\sin^2\cfrac{y}{2}}{\cfrac{y^2}{4}}+\frac{1}{2}L, \end{align*} $

which implies that $L=0$.

Hence,

$ \begin{align*} \lim_{x\to 0} \left(\frac{3x+1}{x}-\frac{1}{\sin x}\right) & = \lim_{x\to 0} \left(3+\frac{1}{x}-\frac{1}{\sin x}\right) = \lim_{x\to 0} \left(3+\frac{\sin x - x}{x \sin x}\right) \\ & = \lim_{x\to 0} \left(3+\frac{\sin x - x}{x^2} \cdot \frac{x}{\sin x} \right) = 3+0 \cdot 1 \\ & = \boxed{3}, \end{align*} $

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1  
This assumes that the limit $L$ exists. How do you know this is true? –  Antonio Vargas Jan 29 '13 at 5:24
    
@Antonio Vargas. You are correct. I also notice that one. Unless $L$ is guaranteed to be finite, then nothing to be worried. –  juniven Jan 29 '13 at 5:27

Here is another approach since we know

$$ \begin{align*} &\lim_{x\to 0} \left(\frac{3x+1}{x} - \frac{1}{\sin x}\right) \\ &\qquad = \lim_{x\to 0} \left(3 +\frac{1}{x} - \frac{1}{\sin x}\right) \\ &\qquad = \lim_{x\to 0} \left(3 + \frac{\sin x - x}{x\sin x}\right) \\ &\qquad = 3 + \lim_{x\to 0} \frac{\sin x - x}{x\sin x} \end{align*} $$

now lets divide the top and bottom by $x^2$ to get

$$3 + \lim_{x\to 0} \frac{(\sin x - x)/x^2}{x\sin x/x^2} = 3 + \lim_{x\to 0} \frac{(\sin x - x)/x^2}{\sin x/x}$$

now using the $lim_{x\to 0} \frac{\sin x}{x} = 1$ we could apply this to the denominator now obtaining

$$= 3 + \lim_{x\to 0} \frac{\sin x - x}{x^2}$$ now lets break the last term up

alright so after we can break this up into

$$\lim_{x\to 0} [(\sin x/x)(1/x) - (1/x) ]$$ now if we factor out $1/x$ we obtain

$$\lim_{x\to 0} (1/x)[(\sin x/x) - 1 ]$$ now lets apply limit property to this to obtain $$\lim_{x\to 0} (1/x)*\lim_{x\to 0}[(\sin x/x) - 1 ]$$

$$\lim_{x\to 0} (1/x)*\lim_{x\to 0}[1 - 1 ] = \lim_{x\to 0} (1/x)*0 = 0$$

so $$3 + 0 = 3$$

:D

now i did not break no calculus rule, sorry its not in latex form still learning a little bit

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5  
@Hamza. Remember that$\frac{\sin x}{x^2}-\frac{1}{x}\neq \frac{\sin x}{x}\left(\frac{1}{x}-\frac{1}{x}\right)$ –  juniven Jan 29 '13 at 4:25
    
@Hamza. The last part of your computations seems not correct to me. –  juniven Jan 29 '13 at 4:33
    
@Derek: Unfortunately, it is not. Look closer... –  Antonio Vargas Jan 29 '13 at 4:37
    
(╯°□°)╯︵ ┻━┻ This is too complicated –  Derek 朕會功夫 Jan 29 '13 at 4:42
    
This is still wrong... $\lim_{x \to 0} 1/x$ doesn't exist, so you can't say that $$\lim_{x\to 0} (1/x)(\sin x/x - 1) = \left(\lim_{x \to 0} 1/x\right)\left(\lim_{x\to 0} (\sin x/x-1)\right).$$ –  Antonio Vargas Jan 29 '13 at 20:12

alright so lets break up the first term to get

lim(x->0) [3x/x + 1/x - 1/sinx] now lets use properties of limits

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) 1/sinx now knowing that lim(x->0) sinx/x = 1 lets divide the third term by x at the numerator as well as the denominator to obtain

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) (1/x)/sinx/x

and since the denominator is just 1 we are left with

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) (1/x)

BUT notice lim x--> 0 1/x does not exist. so this direct method of finding limit would not work. another use is power series as one of the answer is posted :)

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1  
This is wrong for the same reasons that the other answer is wrong. Replace $\sin x$ with $x+x^2$ and try to use the same method to find that limit. It fails; your steps are not valid. –  Antonio Vargas Jan 29 '13 at 3:14
3  
$\lim\limits_{x\to 0}\dfrac{1}{x}$ and $\lim\limits_{x\to 0}\dfrac{1}{\sin x}$ do not exist. –  Jonas Meyer Jan 29 '13 at 3:16
2  
No Hamza, what are you talking about? Try calculating $$\lim_{x \to 0} \frac{3x+1}{x} - \frac{1}{x+x^2}$$ using this method. –  Antonio Vargas Jan 29 '13 at 3:23
3  
The "justifications" you use in each step also apply to that question. If they were valid, they would give you the correct answer there too, but they don't. You get the right answer for the original question on accident. –  Antonio Vargas Jan 29 '13 at 3:27
4  
yeah you are right. my fault. it was a fluke that i got that answer i see what you guys are talking about –  MathGeek Jan 29 '13 at 3:34

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