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Please help me to solve this limit without using L'Hôpital's rule. I don't know what other method can't be used to solve this limit.

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

Martin Sleziak
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Daniel Andrade
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7 Answers7

10

HINT:

$$\frac{1-\cos(x)}{x\cos(x)}=\frac{2\sin^2(x/2)}{x\cos(x)} \tag 1$$

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Using $(1)$ we have $$\begin{align}\lim_{x\to 0}\frac{1-\cos(x)}{x\cos(x)}&=\lim_{x\to 0}\frac{2\sin^2(x/2)}{x\cos(x)}\\\\&=\left(\lim_{x\to 0}\frac{\sin(x/2)}{x/2}\right)\left(\lim_{x\to 0}\frac{\sin(x/2)}{\cos(x)}\right)\\\\&=(1)(0)\\\\&=0\end{align}$$

Mark Viola
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  • Nice! As usual, I assumed the OP knows Taylor Series lol. This is more grounded, may fit better with his knowledge. –  Mar 25 '16 at 16:13
  • @1over137 Thank you very much. Very appreciative! And you are probably correct. -Mark – Mark Viola Mar 25 '16 at 16:16
  • +1 I found this answer very useful, not sure how OP feels. – John Joy Mar 25 '16 at 16:38
  • @johnjoy pleased to hear! And much appreciative of the comment. -Mark – Mark Viola Mar 25 '16 at 17:15
  • Don't you need l'hopitals for $/frac{sin(x)}{x}$ – mike van der naald Mar 25 '16 at 17:16
  • Oh wait, is that where you used Taylor series for sine? – mike van der naald Mar 25 '16 at 17:17
  • @mikevandernaald Recall from elementary geometry that $x\cos(x)\le \sin(x)\le x$. Then, divide by $x$ and apply the squeeze theorem to arrive at the limit in question. L'hospital's Rule is not needed here. -Mark – Mark Viola Mar 25 '16 at 17:29
  • @mikevandernaald See the details of a geometric proof here http://math.stackexchange.com/a/922970/140156 – John Joy Mar 26 '16 at 03:10
  • @mikevandernaald Using LH to compute the limit of $\sin x/x$ at $0$ is a fallacy since LH requires the value of the derivative of the sine at $0$ which by definition is the limit of $(\sin x-0)/(x-0)=\sin x/x$. – Did Aug 22 '16 at 09:02
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This answer is a follow-up to André Nicolas' and Paul Sinclair's comments

$$\lim_{x\to 0}\frac{1-\cos x}{x\cos x}=\lim_{x\to 0}\frac{1-\cos x}{x\cos x}\cdot\frac{1+\cos x}{1+\cos x}=\lim_{x\to 0}\frac{1-\cos^2 x}{x\cos x(1+\cos x)}=\lim_{x\to 0}\frac{\sin^2 x}{x\cos x(1+\cos x)}$$

Note that $\lim_{x\to 0}\frac{\sin x}{x}=1$, that $\lim_{x\to 0}\sin x=0$, that $\lim_{x\to 0}\cos x=1$, and that $\lim_{x\to 0}(1+\cos x)=2$ giving the final result $$\lim_{x\to 0}\frac{1-\cos x}{x\cos x}=0$$

John Joy
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3

Use Taylor series

$$\cos(x) \approx 1 - \frac{x^2}{2}$$ thence

$$\frac{1 - \cos(x)}{x\cos(x)} = \frac{1 - \left(1 - \frac{x^2}{2}\right)}{x\cdot\left(1 - \frac{x^2}{2}\right)} = \frac{x^2}{2x - x^3} = \frac{x^2}{x^2\left(\frac{2}{x} - x\right)} = \frac{1}{\frac{2}{x}} = \frac{x}{2}$$

And since $x\to 0$ the limit is

$$\boxed{0}$$

  • +1 Although there are those on this site who attach some equivalence with LHR and Taylor series. – Mark Viola Mar 25 '16 at 17:35
  • @Dr.MV Super thank you! Maybe Taylor Series can be inappropriate sometimes, but I thought in that case it held ^^ –  Mar 25 '16 at 17:46
  • Yes, absolutely. I actually prefer asymptotics over LHR for many situations. – Mark Viola Mar 25 '16 at 18:47
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Note that $ \displaystyle \lim_{x\to0}\dfrac{1-\cos x}{x} =-\lim_{x\to0}\dfrac{\cos x-\cos 0}{x - 0} =-\left.\cos'x\right|_{x=0} = \sin 0 = 0 $

So $ \displaystyle \lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} = \lim_{x\to 0}\frac{1-\cos(x)}{x} \cdot \lim_{x\to 0}\dfrac{1}{\cos(x)} = 0 \cdot 1 = 0 $

Steven Alexis Gregory
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$\frac{1-\cos x}{x\cos x} = \frac{2 \sin^2 (x/2)}{x \cos x}= \frac{\sin (x/2)}{x/2} \frac{\sin (x/2)}{\cos x}$ The limit is $0$.

imranfat
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kmitov
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0

frist :

$$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\lim_{x \to 0} \frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{2}$$

now :

$$ \lim_{x \to 0} \frac{1-\cos x}{x\cos x}=\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x^2}{x\cos x}\\=\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x}{\cos x}=?$$

since:

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

so :

$$\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x}{\cos x}=(\frac{1}{2})(0)=0$$

Almot1960
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While this question has been answered before, I’d like to offer a more step-by-step solution. This approach may be especially helpful for those who are just beginning to learn about limits.

The question: $$\lim_{x \to 0} \frac{1 - \cos(x)}{x \cos(x)}$$

To solve we can break up the fraction: $$\lim_{x \to 0} \frac {1} {cos(x)} \cdot \frac {1-cos(x)} {x}$$

According to the rules for calculating limits of combinations: $$\lim_{x \to c} {f(x) \cdot g(x)} = \lim_{x \to c} {f(x)} \cdot \lim_{x \to c} {g(x)}$$

In our case $f(x) = \frac {1} {cos(x)}$ and $g(x) = \frac {1-cos(x)} {x}$

So therefore: $$\lim_{x \to 0} \frac {1} {cos(x)} \cdot \lim_{x \to 0} \frac {1-cos(x)} {x}$$

Now we can evaluate each limit separately: $$\lim_{x \to 0} \frac {1} {cos(x)} = \frac {1} {cos(0)} = \frac {1} {1} = 1 $$ and $\lim_{x \to 0} \frac {1-cos(x)} {x} = 0$ is a standard limit.

Since we have found each of our limits, we can now combine them:

$${1} \cdot {0}=0$$

Therefore: $$\lim_{x \to 0} \frac{1 - \cos(x)}{x \cos(x)}=0$$

Nick Pieper
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