$$\lim_{x\to 0}\frac{\sin x}{x}$$

Can't we say this:

$$\lim_{x\to 0}\left[\frac{1}{x}\right]\cdot\lim_{x\to 0}[\sin x]$$

So it would be infinity times 0, which would be 0.

Why is this method wrong?

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    $\begingroup$ Because $\infty \cdot 0$ isn't $0$. $\endgroup$ – MathematicsStudent1122 Nov 27 '16 at 8:45
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    $\begingroup$ I don't think any of the comments and answers so far deal with the actual issue here. The limit of the product equaling the product of the limits is something that is only true when every limit involved exists. The error is right at the second line of your question. $\endgroup$ – Git Gud Nov 27 '16 at 9:00
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    $\begingroup$ Why are questions like this one downvoted? There's nothing wrong with someone wanting an explanation $\endgroup$ – qwr Nov 27 '16 at 9:07
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    $\begingroup$ @GitGud Exactly my thoughts, hence the answer. $\endgroup$ – gowrath Nov 27 '16 at 9:09
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    $\begingroup$ Also, it isn't even true that $\lim_{x\to 0}\frac{1}{x}=\infty$. We have $\lim_{x\to 0+0}\frac{1}{x}=+\infty$ and $\lim_{x\to 0-0}\frac{1}{x}=-\infty$. $\endgroup$ – celtschk Nov 27 '16 at 9:24

Because the rule that you are using, that:

$$\lim a_n b_n = \lim a_n \lim b_n$$

only works if the limits exist. For example here is a screenshot straight from the wikipedia page:

limit laws

Notice how it says, "provided the limits on the right side of the equation exist". In your example, $$\lim_{x \to 0}{1 \over x}$$ does not exist so you can't simply apply the limit laws. Something more refined like l'Hôpital's rule or a more detailed analysis is required (depending on how you defined the trig functions/proved l'Hôptial's).

If you would like a bit of a visual intuition as to why $$\lim_{x \to 0} {\sin x \over x}= 1$$

here is a jank gif I made to show it. I believe one of the other answers discusses this analytically but notice how when you zoom in enough at 0, the graphs of $\sin x$ and $x$ effectively become the same? This is why the ratio of their limits as they approach $0$ goes to $1$.

                                               enter image description here

Maybe as an exercise, compare this to how $\cos(x) - 1$ looks near $x = 0$ and then try and hypothesize an answer to $$\lim_{x \to 0} {\cos x - 1 \over x}$$

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  • $\begingroup$ THANK YOU SO MUCH :o $\endgroup$ – Ahmad Aueda Nov 28 '16 at 15:16

If you accept that any number times $0$ is zero, you must likewise accept that any number times $\infty$ is infinity ! So $0\cdot\infty$ is problematic.

One can illustrate this issue more clearly with

$$1=\lim_{x\to0}\frac xx\stackrel?=\lim_{x\to0}x\cdot\lim_{x\to0}\frac1x.$$

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The expression $\infty \cdot 0$ cannot be interpreted as $0$. Let's think more carefully about what's going on here. We are considering the function $$f(x) = \frac{\sin x}{x}$$ near $x = 0$. Near $x = 0$, the numerator is getting very small. This would suggest that $f(x)$ goes to $0$. But near $x = 0$, the denominator is also getting very small, which would suggest $f(x)$ goes to $\infty$. So the numerator and denominator are competing with each other. The limit thus depends on how which of the two goes to $0$ "faster".

If you think about the graph of $\sin x$, near $x = 0$, it looks very much like the line $y = x$. More precisely, near $x = 0$, $\sin x$ has the linear approximation $$\sin x \approx \sin(0) + x\frac{d}{dx}_{x = 0}\sin x = x.$$ Thus near $x = 0$, $$f(x) \approx \frac{x}{x} = 1,$$ which is why the limit is $1$ and not $0$.

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This limit has $\frac{0}{0}$, so L'Hopital's rule is admissible. Taking the derivative of both the top and bottom, we get $$\lim_{x \to 0} \ \cos(x) = 1$$ In essence, when we see a case of $\frac{0}{0}$ or $\frac{\infty}{\infty} $, it is equivalent to take the derivative of both the numerator and denominator to guage the relative speed at which the two respectively approach those values. If you would like to see an argument that does not involve calculus, you can easily look up a geometric proof using the squeeze theorem on Google.

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    $\begingroup$ Be careful. This may be circular. I won't downvote, though, since this ultimately depends on your definition of $\sin$ $\endgroup$ – MathematicsStudent1122 Nov 27 '16 at 9:02
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    $\begingroup$ For people who are down voting, it would be nice to offer some feedback to the poster of the answer. $\endgroup$ – gowrath Nov 27 '16 at 9:12
  • $\begingroup$ I don't know why whoever down voted did so, but this answer doesn't actually answer the question "Why is this method wrong?", it doesn't even try to do so. The same is true for this answer which also got a (single) down vote. $\endgroup$ – Git Gud Nov 27 '16 at 9:28

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