Can we obtain the following limit structure?

Question

Let us consider we have the following equality in our hand as in the region $(a,b)$ in $ℝ$ $$\tag{1}f(x)=g(x).h(x)$$

1-) The functions $f(x)$ and $g(x)$ are continuous and their results finite (It means we can not see their results as infinity)

2-) And we know that we have the following features:

$$\tag{2}f(x_0)=0 ⇔ g(x_0)=0$$

3-) And the function $h(x)$ is continuous in the region $(a,b)$ . And its all results are finite and non-zero.

Thus, can we obtain the following one from the above-equation (1) with a limit approach with any $x_c$ value in the region $(a,b)$ ?

$$\tag{3}\lim_{x\to x_c}\frac {{f(x)}}{ { g(x)}}=h(x_c)$$

Notice $x_c$ include the zeros $x_0$ ; it means $x_0⊂x_c$.

EXAMPLE: Let me give an example in the region $(-π/2,π/2)$ as :

$$\sin(2x)=2\sin x\cdot \cos x$$ Here

Thus, we can write without any problem:

$$\tag{4}\lim_{x\to x_c}\frac {{\sin(2x)}}{ { \sin x}}=2\cos(x_c)$$

And we know that for $x_0=0$ and $x_0⊂x_c$.

$\sin(2x_0)=0$ and $\sin(x_0)=0$

If $f=g=0$ then we can choose $h$ however we want so I don't know how the last line would make sense in that case — Maximilian Janisch, Sep 29 '19 at 09:51
If $f=g$ is the function that is $0$ in all rational numbers and $1$ in all irrational numbers; and $h=1$ then I also don't know how that makes sense — Maximilian Janisch, Sep 29 '19 at 09:55
Please note $f(x)$ and $g(x)$ are continuous in $(a,b)$. Please see my example $sin2x=2sinx.cosx$ in the question. — MATHH, Sep 29 '19 at 09:56
In that case the section "Quotient rule for limits" here should answer your question — Maximilian Janisch, Sep 29 '19 at 09:59
Please see it again now. I f we repeat, can we obtain the limit (3) without any problem? — MATHH, Sep 29 '19 at 10:16

user · Answer 1 · 2019-09-29T10:19:25.987

1

It is true if we consider the punctured neighborhood of $x_c$ exists for which the ratio $f(x)/g(x)$ is defined.

We can consider pathological cases, as for example the trivial $f(x)=g(x)=0$.

Refer also to the related:

edited Sep 29 '19 at 10:19

answered Sep 29 '19 at 09:59

user

Please see it again now. Soory, ı have not understood your comment. Can we obtain the limit $(3)$ without any problem? – MATHH Sep 29 '19 at 10:17
@GUNES Yes of course in the case you give its ok! – user Sep 29 '19 at 10:18
Thank you. Can you give some more details why we can write equation $(3)$ ? Is it possible for you ? – MATHH Sep 29 '19 at 10:26
@GUNES When the ratio $f(x)/g(x)$ is well defined in the punctured neighborhood of $x_c$ we can take $$\lim_{x\to x_c}\frac {{f(x)}}{ { g(x)}}=\lim_{x\to x_c}\frac {{g(x)h(x)}}{ { g(x)}}=\lim_{x\to x_c} h(x)=h(x_c)$$ since we can cancel out the $g(x)$ factor. – user Sep 29 '19 at 10:29
Thank you, it is good – MATHH Sep 29 '19 at 10:59
You are welcome! Bye – user Sep 29 '19 at 12:00

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