Solve without using L'Hôpital's rule?

Question

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.

Answer 1

Via power series

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

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

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

Answer 2

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 $|x-\sin x|=o(x^2)$.

Answer 3

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, $$ \begin{align} \left|\,\frac{1}{x}-\frac{1}{\sin(x)}\,\right| &=\left|\,\frac{\sin(x)-x}{x\sin(x)}\,\right|\\ &\le\left|\,\frac{\sin(x)-\tan(x)}{x\sin(x)}\,\right|\\ &=\left|\,\frac{1-\sec(x)}{x}\,\right|\\ &=\left|\,\frac{1-\sec(x)}{x}\frac{1+\sec(x)}{1+\sec(x)}\,\right|\\ &=\left|\,\frac{1-\sec^2(x)}{x(1+\sec(x))}\,\right|\\ &=\left|\,\frac{-\tan^2(x)}{x(1+\sec(x))}\,\right|\\ &=\left|\,\frac{\tan(x)}{x}\frac{\tan(x)}{1+\sec(x)}\,\right|\\ &=\left|\,\frac{\tan(x)}{x}\tan(x/2)\,\right| \end{align} $$

Answer 4

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*} $

Answer 5

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

Answer 6

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 :)