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.
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4A start: Multiply top and bottom by $1+\cos x$. – André Nicolas Mar 25 '16 at 16:06
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4Usually $\lim_{x\to 0}\frac{1-\cos x}x = 0$ is proven at the same time as $\lim_{x\to 0} \frac{\sin x}x = 1$ is, in the development of the derivatives of $\sin$ and $\cos$. Since $\lim_{x\to 0} \frac 1{\cos x} = 1$ by continuity, your limit is $0\cdot 1 = 0$. – Paul Sinclair Mar 25 '16 at 16:13
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The answers given here might be useful, too: Evaluating $\lim_{x\to0}\frac{1-\cos(x)}{x}$ – Martin Sleziak Apr 28 '18 at 17:57
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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. – Enrico M. Mar 25 '16 at 16:13
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@1over137 Thank you very much. Very appreciative! And you are probably correct. -Mark – Mark Viola Mar 25 '16 at 16:16
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@johnjoy pleased to hear! And much appreciative of the comment. -Mark – Mark Viola Mar 25 '16 at 17:15
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@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
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@mikevandernaald See the details of a geometric proof here http://math.stackexchange.com/a/922970/140156 – John Joy Mar 26 '16 at 03:10
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@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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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
$$\boxed{0}$$
Enrico M.
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+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
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@Dr.MV Super thank you! Maybe Taylor Series can be inappropriate sometimes, but I thought in that case it held ^^ – Enrico M. Mar 25 '16 at 17:46
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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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@EugenCovaci - If you're at the point where you know what L'Hôpital's rule is, you should already know. – Steven Alexis Gregory Mar 25 '16 at 17:08
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The question is: how do you prove $cos'(x) = - sin(x)$ without L'Hôpital's rule? – Mar 25 '16 at 17:12
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"Please help me to solve this limit without using L'Hôpital's rule" I suppose this means avoinding direct or indirect use of L'Hôpital's rule. – Mar 25 '16 at 17:19
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@EugenCovaci - I didn't use L'Hôpital's rule, I used the definition of the derivative of the cosine function. – Steven Alexis Gregory Mar 25 '16 at 17:22
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@eugencovaci Recall from elementary geometry that $x\cos(x)\le \sin(x)\le x$. Divide by $x$ and apply the squeeze theorem to arrive at the limit in question. No L'Hospital's Rule and no differential calculus was used. – Mark Viola Mar 25 '16 at 17:39
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@eugencovaci I don't believe that we need to start with a result from elementary geometry. It does beg the question as to appropriate assumptions on a fair starting point. – Mark Viola Mar 25 '16 at 18:49
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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 $0$.
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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