Solving the Continuous Function | Hello

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The discussion focuses on the properties of a continuous function f(x) defined over the interval 0 ≤ x < ∞, specifically under the condition that \[\lim_{x\to\infty }f(\frac{1}{ln(x)})=0\]. The correct conclusions drawn from this limit include that f(0) = 0 and f(∞) = 0, confirming that the function approaches zero as its argument approaches infinity. The discussion emphasizes the importance of continuity in evaluating limits and provides a hint regarding the behavior of logarithmic functions at infinity.

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Yankel
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Hello

I need some help with this question, I don't know where to start...

The function f(x) is continuous over 0<=x<infinity and satisfy:

\[\lim_{x\to\infty }f(\frac{1}{ln(x)})=0\]

which conclusion is correct:

1. f(x)=1/ln x

2. f(x)=x

3. f(0)=0

4. f(infinity)=0

5. f(1) = infinity

thanks !
 
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Yankel said:
Hello

I need some help with this question, I don't know where to start...

The function f(x) is continuous over 0<=x<infinity and satisfy:

\[\lim_{x\to\infty }f(\frac{1}{ln(x)})=0\]

which conclusion is correct:

1. f(x)=1/ln x

2. f(x)=x

3. f(0)=0

4. f(infinity)=0

5. f(1) = infinity

thanks !

Hi Yankel!

Uhhhm... I don't know...
Do you have a candidate?
And perhaps a reason to select that candidate?
 
Hint: Since $\lim_{x\to\infty}\ln x=\infty$, it is the case that $\lim_{x\to\infty}(g(\ln x))=\lim_{x\to\infty}g(x)$. Also, $\lim_{x\to+\infty}g(1/x)=\lim_{x\to+0}g(x)$.
 
Easy helpful fact: limits slip past continuous functions. More exactly, if f is continuous and limit g(x) as x approaches a exists, then lim f(g(x))=f(lim(g(x)) -- here a can be either finite or infinite.

Application: 0=lim(f(1/ln x)=f(lim(1/ln x))=f(0)

If you're interested, here's an epsilon delta proof:

View attachment 587
 

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