Finding limits for $\frac{1 - \cos(x)}{x}$

Question

I have some difficulties understanding the last step of the screenshot.

The first step is to multiple the conjugate of the numerator. The second step is to use $1 - \cos^2(x)$ to get $\sin^2(x)$.

How and why did the author splits the $\dfrac{\sin^2x}{x(1+\cos x)}$ into a limit with two functions in it.

Can anyone help to explain the intentions and how it is derived?

Just split the function in the 3rd limit into two function, nothing changed — Tony Ma, Mar 04 '18 at 05:15
The splitting happens because $\sin x\over x$ has a known value as $x\to 0$. Then the second fraction can be evaluated on its own, and it is easy to see that its value is $0$, giving the required result... — abiessu, Mar 04 '18 at 05:16
If $\lim_{x\to a^S} f(x)=L_1\in\mathbb R$ and $\lim_{x\to a^S}g(x)=L_2\in\mathbb R$ then $\lim_{x\to a^S} (fg)(x)=L_1 L_2$. Your case is $lim_{x\to 0}$, so that applies — Tony Ma, Mar 04 '18 at 05:17
A similar asking for a purely geometric argument: Geometric intuition for limit of $\frac{1-\cos x}{x}$ as $x \to 0$ — Silverfish, Oct 03 '25 at 15:18

score 2 · Accepted Answer · answered Mar 04 '18 at 05:19

2

It is known that $\lim_{x \to 0} \frac{\sin x}{x}=1$ and taking that term out make things clearer as now when we substitue $0$ into the denominator, we no longer get $0$ but the numerator still give us $0$.

Since $\lim_{x \to 0} \frac{\sin x}{1+\cos x}=0$, the overall limit is $0$.

answered Mar 04 '18 at 05:19

Siong Thye Goh

155,038

can I treat it as a algebraic equation (split the function in the 3rd limit into two function) where sin(x) is a^2 and x is c, 1+cos(x) is d. i.e. a^2/cd and splits it into a/c * a/d? – ilovetolearn Mar 15 '18 at 13:41
1

yes, but be careful that when we split the function, the limit should exists for each subterm. – Siong Thye Goh Mar 15 '18 at 16:27
thanks for the clarification. – ilovetolearn Mar 15 '18 at 16:27

score 1 · Answer 2 · answered Mar 04 '18 at 05:46

Adding to Siong Thye Goh's answer:

Remember that "the limit of a product is the product of the limits", when each individual factor's limit exists, so that

$$ \lim_{x \to 0} \frac{\sin(x)}{x} \frac{\sin(x)}{1+\cos(x)}$$

$$ = \lim_{x \to 0} \frac{\sin(x)}{x} \lim_{x \to 0} \frac{\sin(x)}{1+\cos(x)}$$

$$ = 1 \times 0 = 0.$$

The key point here is understanding why $\frac{\sin(x)}{x}$ tends to 1, as $x$ tends to 0, and then applying this limit; there are some great geometric proofs of this, if you just do a search here on this site. (Note: I don't recommend that you verify this limit with l'hospital's rule, since it's very much circular reasoning.)

can I treat it as a algebraic equation (split the function in the 3rd limit into two function) where sin(x) is a^2 and x is c, 1+cos(x) is d. i.e. a^2/cd and splits it into a/c * a/d? — ilovetolearn, Mar 11 '18 at 12:51

Finding limits for $\frac{1 - \cos(x)}{x}$

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