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We know that $$\lim_{x \to 0} x \sin \left(\frac{1}{x} \right) = 0$$ When $x \to 0$ then $\sin(1/x)$ is undefined and multiplying this by $0$, the ultimate result is $0$. Why is this not the case when we multiply $0$ and $\infty$?

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    $\lim_{x \to 0} \sin \frac 1x$ is not infinity. Remember that range of sine function on $\Bbb R$ is $[-1,1].$ – 2017-01-11
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    $\infty$ is not a number! – 2017-01-11
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    @VikrantDesai, your comment shows a misunderstanding of the OP's question. – 2017-01-11
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    Because infinity is the inverse of zero, $\infty=1/0$. Then the product must be one, $\infty\cdot0=1/0\cdot0=1$. Infinity is also the square of infinity, $\infty\cdot\infty=\infty$. Then $\infty\cdot0=\infty\cdot\infty\cdot0=\infty\cdot1=\infty$. But the square of $0$ is $0$, so $\infty\cdot0=\infty\cdot0\cdot0=1\cdot0=0$... Morality: the usual computation rules do not apply to infinity. – 2017-01-11
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    Check the discussion and list of _indeterminate forms_ at https://en.wikipedia.org/wiki/Indeterminate_form – 2017-01-11
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    @AlexSilva - $\infty$ IS a number. It is not a *real* (or *complex*) number. However, there are many more sets for which the term "number" is deservedly applied. $\infty$ is an extended real number. It is a number on the Riemann sphere. By identification with $\omega$ or $\aleph_0$, it can be considered an ordinal, cardinal, or surreal number. – 2017-01-11
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    @AlexM. I thought OP wasn't sure about limit of $\sin \frac 1x$ as $x \to 0$. English is not my native language so I struggle sometimes. Apologies. – 2017-01-11
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    here lim x->0 {sin(1/x)} * 0 = 0 and sin(1/0) is undefined and we know that infinity is also undefined. hence if we get 0 * sin(1/0) = 0 then why 0 * infinity is not zero? – 2017-01-11
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    Infinity is not undefined tho infinity*0 is.While sin(1/0) is undefined but 0*sin(1/0) is defined because sin is some number between [-1,1] and 0*some number =0 – 2017-01-11
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    This limit is not computed by "zero times undefined". "Zero times undefined" is, in fact, undefined. – 2017-01-11

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The function $f(x)=\sin\left(\frac{1}{x}\right)$ only takes the values $f \in [-1,1]$.

Thus, you are multiplying $0$ by a finite value, and therefore, the limit converges towards zero.

However, since $\infty$ is not finite like $\sin\left(\frac{1}{x}\right)$ is, when multiplied by $0$, the result is undefined.

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In the extended reals, $0 \cdot \infty$ is undefined.

We define arithmetic to leave it undefined for many of the same reasons why $0/0$ is left undefined.

In particular, we cannot continuously extend multiplication to this case. To extend multiplication continuously, the limit $\lim_{(x,y) \to (0,\infty)} xy$ must be defined; i.e. we must get the same result no matter what path we take approaching $(0, \infty)$.

Here's one path that converges to $1$:

$$ \lim_{x \to 0} x \cdot \frac{1}{|x|} $$

Here's another path that converges to $3$:

$$ \lim_{x \to 1} |x^3-1| \cdot |\cot(x-1)| $$

Since we got two paths with different values, the limit doesn't exist; it is impossible to define multiplication in a way that it is continuous at $(0, \infty)$.