Limit, solution in unusual way

Question

I have a problem with solution of this limit: $$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}$$ Of course, it's a very easy to solve, using (twice) L'Hôpital's rule, but I need to find out, how to do this without this rule.

I stuck in this point: $$\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$$ Everything I need to know is how to eliminate $\frac{\sin^2{x}}{x^2}$, because - as my tutor said - I can't simply substitute $1$ for this expression.

Thanks for help.

PS: It's not a homework. My tutor showed this problem as a puzzle and said, that it would be a good exercise to solve this without L'Hôpital's rule.

EDIT: Here is a way I got to the point $\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$:

\begin{align*} \lim_{x->0}{\frac{\tan{x}-x}{x^2}} &= \lim_{x->0}{\frac{(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}-x)\cdot (\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}{x^2(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}}\\ &= \lim_{x->0}{\frac{\frac{\sin^2{x}}{x^2\cos^2{x}}-1}{x(\frac{\sin{x}}{x}\frac{1}{\cos{x}}+1)}}= \lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}} \end{align*}

Answer 1

Here is an attempt at a geometric proof.

(Figure thanks to J.M)

Consider $\triangle ABC$ such that $\angle{BCA} = x$. Let $BC=1$ and so $AB = \tan x$.

Let $BE$ be the arc of radius 1 and angle $x$ drawn with $C$ as the center (note that $E$ is on AC, between $A$ and $D$ and is kind of hidden in the brown region). Note that $CE = 1$.

Now the area of the gray region is $\dfrac{\tan x}{2} - \dfrac{x}{2}$ (area of $\triangle ABC$ - area of the sector $CBE$).

Let $D$ be the perpendicular on the hypotenuse $AC$ from $B$. It can be seen that $CD = \cos x$ and thus distance from $D$ to $C$ is less than distance from $E$ to $C$ (which is $1$).

Thus the area of the gray region is less than the area of $\triangle BAD$ (gray + brown).

Now $AD = \dfrac{\sin^2 x}{\cos x}$ and thus we have that

$$ 0 < \dfrac{\tan x - x}{2} \lt \dfrac{\sin^3 x}{2\cos x}$$

And so

$$ 0 < \dfrac{\tan x - x}{2x^2} \lt \dfrac{\sin^3 x}{2x^2 \cos x}$$

Since we know that $\lim_{x \to 0+} \dfrac{\sin x}{x} = 1$, and that $\dfrac{\tan x - x}{x^2}$ is an odd function, that the limit is $0$, follows.

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Previous answer, which was a feeble attempt at being pedantic:

For a way to find the limit without using more advanced concepts like McLaurin series etc...

Consider

$$\dfrac{\tan(2x) - 2x}{(2x)^2} = \dfrac{ \frac{2\tan x}{1-\tan^2 x} - 2x}{4x^2}$$

$$ = \dfrac{(2\tan x - 2x) + 2x \tan^2 x}{4x^2(1 - \tan^2 x)} = (\dfrac{\tan x - x}{2x^2} + \dfrac{x\tan^2 x}{2x^2}) \dfrac{1}{1-\tan^2 x}$$

Therefore, taking limits as $\displaystyle x \to 0$

$$ L = (\dfrac{L}{2} + 0)\dfrac{1}{1-0}$$

Thus

$$L = 0$$

There is one problem with the above, though. Can you tell what that is?

(Or rather more simply, replace $\displaystyle x$ with $\displaystyle -x$)

Answer 2

The Maclaurin series for $\tan x$ begins with $x$, and there's no $x^2$ term since it's an odd function. Thus $\tan x = x + O(x^3)$, and therefore $(\tan x - x)/x^2 = O(x) \to 0$ as $x\to 0$.

Answer 3

Note that ${\displaystyle {\tan(x) - x \over x^2}}$ is an odd function, so it suffices to show the limit from either side is zero. So we focus on the right limit, and changing $x$ to $\sqrt{x}$ it suffices to show that $$\lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0$$ We apply the mean-value theorem to $f(y) = \tan(\sqrt{y}) - \sqrt{y}$ on $[0,x]$, which we can do since the mean-value theorem only requires that $f(x)$ is differentiable on the interior of the interval. We obtain that there is a $y \in (0,x)$ such that $$f'(y) = {\tan(\sqrt{x}) - \sqrt{x} \over x}$$ But using the chain rule we have $$f'(y) = {\sec^2(\sqrt{y}) - 1 \over 2 \sqrt{y}}$$ $$= {\tan^2(\sqrt{y}) \over \sqrt{y}}$$ Note that we have $$\lim_{y \rightarrow 0^+} {\tan^2(\sqrt{y}) \over \sqrt{y}} = \lim_{y \rightarrow 0^+}\tan(\sqrt{y})\,\,\,\times \,\,\lim_{y \rightarrow 0^+}{\tan(\sqrt{y}) \over \sqrt{y}}$$ $$ = 0*1 = 0$$ Thus we conclude that ${\displaystyle \lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0}$ as needed.

Answer 4

Let's solve it elementarily considering $\sin(x)<x<\tan(x),\space 0< x <\frac{\pi}{2}.$ From this inequality we
get that: $$0\leq\lim_{x\to 0}{\frac{\tan{(x)}-x}{x^2}}\leq\lim_{x\to 0}{\frac{\tan{(x)}-\sin(x)}{x^2}} \tag1$$ $$\lim_{x\to 0}{\frac{\tan{(x)}-\sin(x)}{x^2}}=\lim_{x\to 0}{\frac{\sin(x)(\frac{1}{\cos(x)}-1)}{x^2}}=\lim_{x\to 0} \frac{\sin(x)}{x} \lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)}=$$ $$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)}=\lim_{x\to 0} \frac{1-\cos^2(x)}{x\cos(x)(1+\cos(x))}=\lim_{x\to 0} \frac{x\sin^2(x)}{x^2\cos(x)(1+\cos(x))}=0.\tag2$$

From $(1)$ and $(2)$ by applying Squeeze's theorem we get that $L=0$.

The proof is complete.

Answer 5

I am not sure what you can use; if you know of Maclaurin series, the argument is very easy.

I believe that what your tutor meant is that you cannot rewrite $\displaystyle \frac{\tan x-x}{x^2}=\frac{\frac{\sin x}x-\cos x}{x\cos x}$ as $\displaystyle \frac{1-\cos x}{x\cos x}$. The argument below uses that $\sin x/x\to 1$, but not by substituting it this way.

Note that $\displaystyle \frac{1-\cos x}{x^2}=\frac{1-\cos^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}x\right)^2\frac1{1+\cos x}\to\frac12$.

It is enough to find the limit of $\displaystyle \frac{\frac{\sin x}x-\cos x}x$. Using the limit above, all you need is to find the limit of $\displaystyle\frac{\sin x -x}{x^2}$, because $$\frac{\frac{\sin x}x-\cos x}x=\frac{\frac{\sin x}x+\left(\frac{1-\cos x}{x^2}\right)x^2-1}x.$$

The limit of $\displaystyle\frac{\sin x -x}{x^2}$ is 0. This is fairly easy to evaluate using the Maclaurin expansion of $\sin x$. All you really need is that $\sin x=x+O(x^3)$, but I am not sure you are familiar with this notation. Also, you could define $f(x)=\sin x/x$ if $x\ne 0$ and $f(0)=1$, and verify that this function is differentiable at 0. But I am not certain you have the tools to do that.

Answer 6

My inclination is to rewrite the original limit as:$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}=\lim_{x\to 0}{\frac{\frac{\tan{x}}{x}-1}{x-0}},$$ with the thinking that if $f(x)=\frac{\tan x}{x}$, $\lim_{x\to 0}f(x)=1$ and $\lim_{x\to 0}f'(x)=0$ and $$f'(a)=\lim_{x\to a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\to a}{\frac{\frac{\tan x}{x}-\frac{\tan a}{a}}{x-a}}.$$ Unfortunately, to proceed from there, I'm getting stuck in some ugliness with double limits.

Answer 7

$lim_{x \to 0} \frac{tan(x)−x}{x*x}$ (0/0 Form)

$lim_{x \to 0} \frac{sec(x) * sec(x) -1}{2x}$ (0/0) Form $\frac{d(Numerator)}{d(Denominator)}$

$lim_{x \to 0} \frac{2secx * sec(x) *tan(x)}{2} = sec0 * sec0 *tan0 =1 * 1 * 0 =0$

Ans is 0.