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I want to know when can I replace $\sin(x)$ by $x$ when $x$ approaches zero. My teacher taught me that when $x$ approaches zero, $\sin(x)/x = 1$. Now, in large expressions of the form $\sin(x)\cdot f(x)$ or $\sin(x) + f(x)$, when can I replace $\sin(x)$ by $x$.

If possible, please provide me more material on where can I learn about these shortcuts.

Jimmy R.
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Groverkss
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  • I think it is important for you to understand what means "when x approaches zero, sin(x)/x=1" – MysteryGuy Apr 03 '18 at 09:22
  • You can't replace $A$ by $B$ unless $A=B$. Thus replacing $\sin x$ by $x$ is not allowed. But whenever you see the expression $\lim_{x\to 0}\dfrac{\sin x} {x} $ then you can replace it by $1$ precisely and simply because that expression is equal to $1$. – Paramanand Singh Apr 03 '18 at 14:37
  • @user516256 Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 07 '18 at 19:49

4 Answers4

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You can always do this:

$$\lim_{x\rightarrow 0} (\sin{(x)} \cdot f(x))= \lim_{x\rightarrow 0} (\frac{\sin{(x)}}{x}\cdot x \cdot f(x))$$ If $$\lim_{x\rightarrow 0} ( x \cdot f(x)) = a \Rightarrow \lim_{x\rightarrow 0} (\sin{(x)} \cdot f(x))= \lim_{x\rightarrow 0} (\frac{\sin{(x)}}{x}\cdot x \cdot f(x))= 1\cdot a = a$$

trancelocation
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If $f\sim g$, i.e. $\lim\limits_{x\to somewhere}\frac fg=1$ then for any $\alpha\in \mathbb R$ we have $ f^\alpha h= \left(\frac{f}{g}\right)^\alpha\cdot g^\alpha h$ and so $ \lim f^\alpha h= \lim g^\alpha h$ whenever one of the limits exists.

So we can replase $f$ by $g$ in such expressions without changing the result. Such replacement in the sums there is no justification except the desire and may lead to wrong results.

Let $f=x+1,\,\, g=x$ then $f\sim g$ when $x\to \infty$. Try to replase $f$ by $g$ in the $\lim\limits_{x\to\infty} (f-g).$ :)

Minz
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No it is not possible in general make this kind of substitution without take the all limit at once. We only can when the first order term is the one which really determine the limit and in any case always using little-o or Big-O notation in order to handle properly the remainders for higher order terms.

Take also a look here for an example and some references Problem with limit solving.

user
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  • The thing is, I was told by my teacher that it is possible if in multiplication. Now I want to explore about it and check if it's right. – Groverkss Apr 03 '18 at 09:16
  • Of course we can't in general for $\sin(x) + f(x)$ see the example in the link. – user Apr 03 '18 at 09:20
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For small $x$, $\sin x$ is indeed "of order" $x$. So you can replace it by $x$ unless it is canceled by another term of order $x$ (when subtracted).

For instance, $e^x-1$ is also of order $x$ and

$$\frac{\sin x}{e^x-1}\to\frac xx=1$$ is valid.

And

$$\sin x-(e^x-1)\to x-x=0$$ is not because the cancellation exposes residual lower order terms.