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I've just discovered something really interesting (Or to be precise, it's been discovered for ages but I've just realized it now.)

We know these:

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

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

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

However, can we generalize it like this:

For all polynomial $f(x)$ that has roots $a_i$, where $a_i\in \Bbb{Z}$, then

$$\lim_{x\to a_i} \frac{\tan{f(x)}}{f(x)} = 1$$,

$$\lim_{x\to a_i} \frac{\sin{f(x)}}{f(x)} = 1$$, and

$$\lim_{x\to a_i} \frac{1 - \cos{f(x)}}{f(x)} = 0$$

I'm not going to write all functions that I've been tried, but here's one of them:

$f(x)=x^2+x-6$ has roots $-3$ and $2$.

That means,

$$\lim_{x\to -3} \frac{\tan{(x^2+x-6)}}{x^2+x-6} = 1$$

and

$$\lim_{x\to 2} \frac{\tan{(x^2+x-6)}}{x^2+x-6} = 1$$

and perhaps it holds for the other related limits. Nevertheless, how to prove it if this is true (please, give a counterexample if it isn't)? Also, what if $a_i\in \mathbb{R}$? Thanks in advance!

user516076
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1 Answers1

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Whenever you have $\lim_{x\to0}f(x)=l$ and $a$ is a zero of a continuous function $g$, then $\lim_{x\to a}g(x)=g(a)=0$, and therefore $\lim_{x\to a}f\bigl(g(x)\bigr)=l$. This explains to all of your limits.

José Carlos Santos
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