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I'm trying to prove that if $\lim_{x\to 0^+} xf(x) = L > 0 $ then $\lim_{x\to 0^+} f(x) = \infty$

I'm wondering if the limit laws for the sum of functions can be applied here. Can I write $$\lim_{x\to 0^+} \left(xf(x) - L\right) = 0 \\ \implies \lim_{x\to 0^+} \left(f(x) - \frac{L}{x}\right) = 0 \\ \implies \lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} \frac{L}{x} = \infty$$

Or do I need to do a proof using the $\delta-M$ definition of the limit? In the definitions for the limit laws it usually starts with the fact that that limits exist for both functions but in this case one of them diverges.

Thanks

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Hint. Recall that, when both sides are meaningful, you have $$ \lim_{x\to 0} \left(\frac{u(x) }{v(x)}\right)= \frac{\displaystyle \lim_{x \to 0}u(x) }{\displaystyle \lim_{x \to 0}v(x)} $$ then, for $x$ tending to $0$, $x \neq 0$, write $$ f(x)=\frac{xf(x) }{x}. $$

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