Log 10 (subscript) x - Limit

[Jason76]

Log 10 (10 is in subscript) x

The book says it is undefined at x = 0 and is also discontinuous there (and the book says it has no limit). How does that differ from the limit of $\displaystyle \dfrac{1}{x^{2}}$ as x approaches 0 (the limit is positive infinity) and the limit of $\displaystyle \dfrac{1}{x^{2}}$ as x approaches positive infinity (limit is 0)?

 

[daon2]

Log 10 (10 is in subscript) x

The book says it is undefined at x = 0 and is also discontinuous there (and the book says it has no limit). How does that differ from the limit of $\displaystyle \dfrac{1}{x^{2}}$ as x approaches 0 (the limit is positive infinity) and the limit of $\displaystyle \dfrac{1}{x^{2}}$ as x approaches positive infinity (limit is 0)?

The book is probably adhering to its definition of a limit existing. Most go like this:

$\displaystyle \displaystyle \lim_{x\to a} f(x) = L \iff \lim_{x\to a^{+}} f(x) = L = \lim_{x\to a^{-}} f(x)$

So a left and right inverse must exist and be equal. We cannot talk about the left-hand limit $\displaystyle \displaystyle \lim_{x\to 0^{-}} \log_{10}(x)$ since $\displaystyle \log_{10}(x)$ is not defined for $\displaystyle x<0$. However it is okay to say

$\displaystyle \displaystyle \lim_{x\to 0^{+}} \log_{10}(x) = -\infty$

 

[pka]

Log 10 (10 is in subscript) x
The book says it is undefined at x = 0 and is also discontinuous there (and the book says it has no limit). How does that differ from the limit of $\displaystyle \dfrac{1}{x^{2}}$ as x approaches 0 (the limit is positive infinity) and the limit of $\displaystyle \dfrac{1}{x^{2}}$ as x approaches positive infinity (limit is 0)?

I am not sure what the last part of the question means.

But $\displaystyle y=\log_{10}(x)$ is defined if and only if $\displaystyle x>0$.

And $\displaystyle y = \log _{10} (x) \Leftrightarrow x = 10^y$

Note that \(\displaystyle \displaystyle\lim _{y \to - \infty } 10^y = 0
\)

That is why your book says $\displaystyle \displaystyle\lim _{x \to 0^ + } \log _{10} (x)$ is not defined.

 

[HallsofIvy]

Since "infinite" is not a member of the real number system, saying that "the limit is infinity" just says that the limit does not exist for a particular reason. If we take the limit, as x approaches 0, of $\displaystyle 1/x^2$, from both sides, the value of $\displaystyle 1/x^2$ is a large positive number. The limit does NOT in fact exist but a standard notation for that is "$\displaystyle \infty$". If the function had been 1/x instead, x close to 0 and positive 1/x would be a large positive number. If x is close to 0 and negative 1/x would be a large negative number. Again, the "limit" does not exist but now we would NOT say "$\displaystyle \lim_{x\to 0} \frac{1}{x}= \infty$". We could say that "$\displaystyle \lim_{x\to 0^+}\frac{1}{x}= \infty$" and $\displaystyle \lim_{x\to 0^-} \frac{1}{x}= -\infty$. With $\displaystyle log_{10}(x)$, the situation is different from both of those. For x positive but close to 0, $\displaystyle log_{10}x$ will be a very large negative number. That is, $\displaystyle \lim_{x\to 0^+} log_{10}(x)= -\infty$. For x negative, $\displaystyle log_{10}(x)$ is not even defined. There cannot be any "limit as x goes to 0 from below".

 

[Jason76]

If log (10 in subscript) x has no limit, then it is divergent. A divergent function comes about when y (dependent variable) goes toward infinity as positive numbers are plugged into x. Is that right? The log function differs from certain fractional functions where x is in the bottom, because fractional functions of that type decrease (go toward 0) as x increases and visa versa.

so

$\displaystyle \dfrac{1}{x^{2}}$ - y decreases (goes toward 0 without reaching it) as x increases, and so has a limit of 0. However, If y increases (goes toward positive infinity), then the x values plugged in are decreasing. On the other hand, with the log (10 in subscript) x, y values increase, but x values also increase, because no fractional inverse relationship exists between numerator and denominator. Positive infinity has no limit. Now if x is decreasing, then y decreases, because, as we stated, this log function is not an "inverse relationship".