As it tends...

Let $f : [0,1] \rightarrow [-1,1]$ be a non-zero function such that

$f(2x)=3f(x), x\in [0, \frac{1}{2}].$

Then $\lim_{x \to 0+} f(x)$ is equal to ?

#Calculus

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I think that the answer is 0
Proof:-
Taking x = 1/2 and 1/4 and 1/8 and 1/(2^n), we get that
[f(1)]/(3^n) = f( 1/(2^n) )
Taking the limits as n approaches infinity, we get the desired answer.........The RHS is what you asked for.......LHS tends to 0
Please tell is I am wrong

Aaghaz Mahajan - 3 years, 2 months ago

f(2x)=3f(x)= $3^2$ f( $\frac{x}{2}$ )=....= $3^{n}$ *f( $\frac{x}{2^{n-1}}$ ).So f( $\frac{x}{2^{n-1}}$ )= $\frac{f(2x)}{3^n}$ ,So as n tends to $\infty$ ,f(0)=0 since f(2x) is finite.

rajdeep brahma - 2 years, 12 months ago

From the discussion above, we have $x \in [0, \frac{1}{2}]$

We can assume that the $\lim_{x\to 0+} f(x) = \lim_{x\to 0+} f(2x) = a$ , where $a$ is the limit $\forall x \in [0, \dfrac{1}{2}] \quad ...(1)$

Now, we have, $f(x) =\dfrac{f(2x)}{3}$

From $(1)$ we have

$\lim_{x\to 0+} f(x) = \lim_{x\to 0+} \dfrac{f(2x)}{3}$

$\implies \lim_{x\to 0+} f(x) =\dfrac{lim_{x\to 0+}f(2x)}{3}$

$\implies a = \dfrac{a}{3}$

$a(3-1)=0$

Since, $2 \neq 0\ \implies a=0$

Dhruva V - 3 years, 2 months ago

OHHHH!!!! I see......A nice method......!! What do u think of my solution?? Is it fine??

Aaghaz Mahajan - 3 years, 2 months ago

It is fine if you intend to find the limit by observation, which is rather an informal way of finding solutions of pre-calculus problems. A definitive proof-aided approach is recommended.

Dhruva V - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$