3

I was going through a Calculus Textbook, when came across the following two identities :

$$\lim \limits_{x \to 0}\dfrac {\sin (x)}{x} = 1 \qquad\text{and}\qquad\lim \limits_{x \to 0}\dfrac {\tan (x)}{x} = 1.$$

The first one is really popular/famous and has many proofs. But I am more interested in the second one.

The fraction $\frac {\tan (x)}{x}$ becomes $\frac {0}{0}$ at $x=0$, so the easiest approach is the L'Hopital's Rule.

So $$\lim \limits_{x \to 0}\dfrac {\tan (x)}{x} = \dfrac{\frac{d}{dx}\tan(x)}{\frac{d}{dx}x} = \dfrac{\sec^2(x)}{1} \\ \implies \lim \limits_{x \to 0}\dfrac {\tan (x)}{x} = \sec^2(0) = 1.$$

A geometrical proof of the same can be found here. I am looking for some non-geometrical proof for this identity.

Xander Henderson
  • 29,772
  • 10
    $\frac{\tan x}{x}=\frac{\sin x}{x}\frac{1}{\cos x}$ – egreg Feb 12 '17 at 15:44
  • 2
    If you know that the derivative of $\tan\theta$ is $\sec^2\theta$, then the definition of the derivative tells you $$\lim_{x\to0}{\tan x\over x}=\lim_{x\to0}{\tan x-\tan0\over x-0}=\sec^2(0)=1$$ without resorting to L'Hopital. – Barry Cipra Dec 15 '19 at 15:46

4 Answers4

7

$$\lim \limits_{x\rightarrow 0} \dfrac{\tan{x}}{x}=\lim \limits_{x\rightarrow 0} \dfrac{\sin{x}}{x}\lim \limits_{x\rightarrow 0} \frac{1}{\cos{x}}=1\times 1=1 $$

since the two limits exist.

Kal S.
  • 3,781
3

Using that

$$\lim_{x \to 0} \frac{\sin x}{x}=1$$

and

$$\lim_{x \to 0} \frac{1}{\cos x}=\frac{1}{\cos 0}=1 $$

We get, using product of limits

$$1=\lim_{x \to 0} \frac{\sin x}{x\cos x}=\lim_{x \to 0}\frac{\frac{\sin x}{\cos x}}{x}=\lim_{x \to 0} \frac{\tan x}{x}$$

Nicolás Vilches
  • 3,525
  • 1
  • 14
  • 14
3

You know $$\lim \limits_{x \to 0}\dfrac {\sin (x)}{x} = 1$$ so $$\lim \limits_{x \to 0}\dfrac {\tan (x)}{x} =\lim \limits_{x \to 0}\dfrac {\sin (x)}{x}\times\frac{1}{\cos(x)}= 1\times\frac{1}{1}=1$$

Nosrati
  • 29,995
3

If you know the proof of $\lim_{x \to 0} \sin x/x = 1$ then you may know about following inequality,

$$\cos(x)\le\frac{\sin(x)}{x}\le1$$

Dividing by $\cos x$,

$$1\le\frac{\tan(x)}{x}\le \sec x$$

Taking the limit,

$$1\le\lim_{x \to 0} \frac{\tan(x)}{x}\le 1$$