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Recently in my calculus class I have been taught $\lim_{x\to 0}\frac{\sin x}{x}$ =1. Now, in trigonometry I have studied $\sin 0=0$. Also, I don't see $\sin x/x=1$. Now my confusion is that if I have to use these identities in some practical applications , which of them should I use?

tanya
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    I think you need to read http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – Juniven Acapulco Jan 11 '17 at 03:48
  • I know the proof of $lim_{x->0}sinx/x=1 but I need to the difference between the three expressions as mentioned in my question – tanya Jan 11 '17 at 03:55
  • From the taylor expansion of $\sin x$ one can see it behaves like $x$ around $0$. So they cancel out. The fraction behaves like $1$ near $0$. – AHB Jan 11 '17 at 03:59

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$\lim_{x \to 0} \frac{\sin x}{x}$ is a expression that evaluates to $1$.

$(\sin x)/x$ is a function of $x$.

$(\sin 0)/0$ is an undefined quotient $= 0/0$.

I won't go into details why $\lim_{x \to 0} \frac{\sin x}{x} = 1$ but you need to remember that the value of $\sin$ at $0$ has nothing to do with this limit.

Henricus V.
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If you know the Taylor expansion of the function around limit point, you can substitute the function with its Taylor expansion around the limit point

This is a good way of saving time in limit computation.

$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!} - \dots$$

$$\implies \sin x = \mathcal{O}(x) \ \ \ \ \ (x \to 0)$$

$$\implies \lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to0}\frac{x}{x}=\lim_{x\to0}{1}=1$$

AHB
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    You need to make precise (with the little-o notation, for example) what you mean by $$\sin x \approx x \ \ (\text{around }0)$$ Otherwise this is not a proof. – Guest Jan 11 '17 at 04:10
  • No proof was intended. Just a short cut was introduced, @Guest . Proof is available in the linked question. – AHB Jan 11 '17 at 04:41
  • @Guest Is it correct now? – AHB Jun 18 '17 at 11:06
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The limit as $x\to0$ of a function has nothing to do with the value of that function at $0$.

If you are working with real-valued functions, then the maximal domain for $f(x):=\frac{\sin x}{x}$ is $\mathbb{R}\backslash\{0\}$ since at $0$ you get a division by $0$ which is undefined. The expression $\frac{\sin0}{0}$ doesn't make sense because of this. However, it is customary to write $\frac{\sin x}{x}$ when we are really considering the sinc function $$ \operatorname{sinc}x:=\begin{cases}\frac{\sin x}{x},&&x\neq0,\\1,&&x=0\end{cases} $$ so it is possible that using this definition, $\frac{\sin 0}{0}$ is meant to equal $1$. Note that this value is the one which extends continuously $f$ at $0$ because $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$.

There are no real numbers for which $\frac{\sin x}{x}=1$. As mentioned earlier, $x=0$ won't work because you get an undefined expression, and you can convince yourself graphically that there are no nonzero $x$ for which $\sin x=x$ (or, better, prove it).

Guest
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As the question is of indeterminate form, by using L'Hospital rule we have $$\lim_{x\to 0} \frac {\sin x}{x}=\lim_{x\to 0} \frac {\cos x}{1}=1$$

Rohan Shinde
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