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Is the following statement correct? If yes, prove it. $$\lim_{x \to 0^+} \log x = - \infty$$

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$\require{enclose}$

  • $\lim_{x \to a}f(x) = L$ means that $f(x)$ can be made as close to $L$ as desired, by making $x$ close enough (but not equal) to $a$.
  • $\lim_{x \to a^\enclose{box}[mathcolor="red"]{\color{black}+}}f(x) = L$ means that $f(x)$ can be made as close to $L$ as desired, by making $x$ close enough (but not equal) to $a$ from the right side on the number line.
  • $\lim_{x \to a^\enclose{box}[mathcolor="red"]{\color{black}+}}f(x) = -\infty$ means that $f(x)$ can be made as largely negative as desired, by making $x$ close enough (but not equal) to $a$ from the right side of the number line.

To Prove: For any given number $M$, there exists a value of $x$ sufficiently close to $0$, such that the value $\log x$ is smaller (more largely negative) than $-M$. That is, $$\log x < -M$$

Exponentiating both sides, we get:$$\begin{align} e^{\log x} &< e^{-M}\\[1em] \implies x &< e^{-M} \end{align}$$

Since $e^{-M} > 0$ for any $M \in \mathbb{R}$, we can choose our $x$ such that $0 < x < e^{-M}$.

Thus, we have shown that no matter how largely negative $-M$ is, $\log x$ can be made more negative than $-M$ by making $x$ sufficiently close to $0^+$.

And that is exactly what $\lim_{x \to 0^+} \log x = -\infty$ means.

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