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Let $f:\left(0,\infty\right)\longrightarrow\mathbb{R}$ be a monotonically increasing function.

Let $g:\left(0,\infty\right)\longrightarrow\mathbb{R}$ , $ g\left(x\right)=\frac{f\left(x\right)}{x}$ is a monotonically decreasing function.

How can I prove that $ f $ is continuous?

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What happens to $g$ at $x$ if the limit of $f(y)$ when $y\to x$, $y<x$, is strictly less than $f(x)$? –  Did Jan 29 '12 at 17:35
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up vote 4 down vote accepted

Since $f$ is a monotonically increasing function, we have for all $a>0$, $$ b=\lim_{x\uparrow a} f(x) \le \lim_{x\downarrow a} f(x)=c. $$ Then $$ \lim_{x\uparrow a} \frac{f(x)}{x} =\frac ba \le \frac ca =\lim_{x\downarrow a} \frac{f(x)}{x}. $$ But if the function $x\mapsto f(x)/x$ is monotonically decreasing then $$ \lim_{x\uparrow a} \frac{f(x)}{x} \ge \lim_{x\downarrow a} \frac{f(x)}{x}. $$ Hence these last two one-sided limits are equal, so $g$ is continuous. If $x\mapsto g(x)$ is continuous, then $x\mapsto xg(x)$, being the product of two continuous functions, is also continuous.

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Why is g continuous if the two one-sided limits are equals? As far as I understand, if the two one-sided limits exists it means that the limit at this point exist but how do we know that indeed limit at a equal to the function value at a? –  sony jimbo Jan 30 '12 at 6:28
    
@sonyjimbo : It was given that $g$ is monotonic. If the limit of a monotonic function exists at a point, and the function is defined at that point, then the value of the function at that point must be equal to the limit, since inequality of those two things would be a lack of monotonicity. –  Michael Hardy May 30 at 15:05
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at each $x\in (0,\infty)$ the one sided limits of $f$ at $x$ must exist and must satisfy the obvious inequality. Do the same for $g$ and conclude that the right and left limits of $f$ must be equal and since $f$ is monotonic, it follows that $f$ is continuous at $x$.

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