Classify the type of discontinuity at $x_0 = 0$

Question

for (a) I think it is essential because the right side goes to infinity. for (b) I think it is removable because the function is not defined in $0,$ same goes for (c)

I am really not sure about these.

Answer 1

(a) $$\lim_{x\to 0^+}x^x = 1$$ (take $\log$)

(b) $$\lim_{x\to 0^+}\frac1x\sin(1/x)\cos(1/x),\qquad\lim_{x\to 0^-}\frac1x\sin(1/x)\cos(1/x)$$ does not exists because the oscillating factor $\sin(1/x)\cos(1/x)$.

(c)

$$\lim_{x\to 0^+}e^{\text{sgn}(x)(|x|+1)} = e^1,$$ $$\lim_{x\to 0^-}e^{\text{sgn}(x)(|x|+1)} = e^{-1},$$

Answer 2

Case (a) For $x > 0$ you have $x^x = e^{x \ln x}$. As $\lim\limits_{x \to 0+} x \ln x = 0$, you have $\lim\limits_{x \to 0+} f(x)=1$. Hence the discontinuity is removable.

Case (b) For $x \neq 0$, $$f(x)=\frac{\sin \frac{2}{x}}{2x}.$$ For $x=\frac{1}{k \pi+\frac{\pi}{4}}$ (with $k$ integer) you have $f(x_k)=\frac{k \pi}{2}+\frac{\pi}{8} \to \infty$ as $k \to \infty$. Therefore $\lim\limits_{x \to 0} f(x)$ doesn't exist and the discontinuity is essential.

Case (c) Here you have a jump as $$\begin{cases} \lim\limits_{x \to 0^-} f(x) = e^{-1}\\ \lim\limits_{x \to 0^+} f(x) = e^{+1} \end{cases}$$

Answer 3

If I remember correctly:

These are all singularities because they are not defined at $x_0 = 0$. But there are three ways $f$ might behave near 0.

1) "removable": the limits, both of them, approach a consistent and equal value. It as though the function just has a "hole" in it. It's called "removable" because it's as though we could just remove the hole and plug it in with the value of the limits.

2) "jump": the limits on one side exist but aren't the same as the limit from the other side. The function "jumps".

3) Essential: The limits don't exist.. It's not simply a "break".

a) $x^x$. $0^0$ isn't defined as it would create inconsistencies. But $lim_{x \rightarrow 0} x^x$ does exist. $lim_{x \rightarrow 0} x^x = e^{x\ln x} = e^x*1 = e^0 = 1$. This is removable.

b) lims of sin(1/x) and cos(1/x) oscillate but don't converge as x goes to 0. 1/x tends to infinity. The limits don't converge so b) is an essential singularity.

c) $\lim_{x \rightarrow 0^+}f(x) = e^{x + 1} = e$

$\lim_{x \rightarrow 0^-}f(x) = e^{x - 1} = 1/e$

These limits both exist but aren't equal. This is a "jump" singularity.