calculus - Is $f$ continuous at zero?

$$\require{cancel}$$

$$f(x) =
\begin{cases}
\frac{\sin x}{|x|} &\text{ if }x \neq0
\\
\hspace{0.3cm}1 &\text{ if }x=0.
\end{cases}$$

My Attempt

1)$$\lim_{x\rightarrow0}\frac{\sin x}{|x|} = \lim_{x\rightarrow0}\frac{\sin x}{x} \frac{x}{|x|} = 1\lim_{x \rightarrow0}\frac{x}{|x|}
\\$$
2)$$\lim_{x\rightarrow0^{-}}\frac{x}{|x|}=-1 \hspace{0.3cm}\text{and}\hspace{0.3cm} \lim_{x\rightarrow0^{+}}\frac{x}{|x|}=1
$$
Therefore:
$$1\lim_{x\rightarrow0}\frac{x}{|x|}=DNE
$$
so, $f$ is not continuous at $0$.

My question is does my solution actually prove that $f$ is not continuous at $0$? or is it continuous at zero because $f(x)=1$ when $x=0$?

Answer

1)$$\lim_{x\rightarrow0}\frac{\sin x}{|x|} = \lim_{x\rightarrow0}\frac{\sin x}{x} \frac{x}{|x|} = 1\lim_{x \rightarrow0}\frac{x}{|x|}
\\$$

Note that the second equality does not hold, since for two sequences $(a_n)$, $(b_n)$ you only have
$$
\lim_{n \to \infty} a_n b_n = \lim_{n \to \infty} a_n \lim_{n \to \infty} b_n
$$

provided that both sequences converge. Hence in your case it would be better to start directly with a modification of part 2) :

If $f(x)$ is continuous at $0$ then the following equality is necessary:
$$
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x).
$$
But on the one hand you have
$$
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin(x)}{|x|} = \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1
$$

and on the other hand
$$
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(x)}{|x|} = \lim_{x \to 0^-} \frac{\sin(x)}{-x} = -\lim_{x \to 0^+} \frac{\sin(x)}{x} = -1.
$$
So $f$ can't be continuous at $0$.

calculus - Monotonic function non-continuous in each rational

How can I prove that exists a monotonic non-decreasing function $f: [0,1] \rightarrow \mathbb R$ that isn't continuous in every rational of its domain?

elementary number theory - How to prove that $53^{103}+ 103^{53}$ is divisible by 39?

This is a problem in my number theory textbook. It is based on modular arithmetic but im not getting how to start off to prove this. Please give me some hints on how to solve it.

Answer

As $39=13\cdot3$

For non-negative integers $m,n$

$\displaystyle53\equiv1\pmod{13}\implies53^n\equiv1$ and $\displaystyle103\equiv-1\pmod{13}\implies103^{53}\equiv(-1)^{53}$

$\displaystyle\implies53^{103}+103^{53}\equiv1+(-1)\pmod{13}$

and $\displaystyle53\equiv-1\pmod3\implies53^{103}\equiv(-1)^{103}$ and $\displaystyle103\equiv1\pmod3\implies103^m\equiv1$

$\displaystyle\implies53^{103}+103^{53}\equiv-1+(1)\pmod3$

logarithms - $sqrt{log n}$ vs $logsqrt{n}$

I have to check if $ \sqrt{ \log(n) } = \theta(\log({\sqrt n}))$.
Following log rules I can write:

$ \sqrt{ \log(n) } = \log(n)^{\frac{1}{2}}$
$\log({\sqrt n)} = \log(n^{\frac{1}{2}}) = \frac{1}{2}\log(n) $

Looking at graphs I can see the $O$ notation is correct but not the $\Omega$.
I would appreciate some help at how to disprove this, as I am stuck for quite some time on it.

Thanks in advance

Answer

You want to prove that its not true that $\sqrt{\log(n)} = \Omega(\log\sqrt n)$. It is enough to prove that $\frac{\log\sqrt n}{\sqrt{\log n}}$ is unbounded. But this is true, simply because this ratio is

$$ \frac{\log\sqrt n}{\sqrt{\log n}} = \frac{\sqrt{\log n}}{2} \to \infty,$$
for $n$ large.

number theory - Solving Linear Congruences.

Question is -:

Solve the linear congruence $3x \equiv 4\left(mod\, \, \, \, 7\right)$, and find the smallest positive integer that is a solution of this congruence

My Approach-:

$3x \equiv 4\left(mod\, \, \, \, 7\right)$

$\Rightarrow x \equiv 3^{-1}\, \,4\left(mod\, \, \, \, 7\right)$

$3^{-1}$ means it is the multiplicative inverse of $3\, \,mod\, \,7$

multiplicative inverse of $3\, \,mod\, \,7$

$\Rightarrow 7=3*2+1$

$\Rightarrow 3=1*3+0$

$\Rightarrow 1=1*7+\left(-2\right)3$

thus $-2$ or $5$ is the inverse.

Thus i am getting

$\Rightarrow x \equiv 3^{-1}\, \,4\left(mod\, \, \, \, 7\right)$

$\Rightarrow x \equiv 20\left(mod\, \, \, \, 7\right)$

But in the solution they are multiplying the inverse $5$ to both sides and get equation as-:

$15 \,x \equiv20 \, \left(mod\,\,7\right) $

and then

$x \equiv 15x\,\equiv\,20\,\equiv\,\,6\,\left(mod\,\,7\right)$

The solution is given here

Please help me out ,where i am wrong!

thanks!

Answer

You're not wrong, when you write $x \equiv 3^{-1} 4$, then you have
also multiplied both sides by the inverse of $3$. You're just using
different notation.

elementary number theory - Prove that if $gcd(a,b)=1$, then $gcd(acdot b,c) = gcd(a,c)cdot gcd(b,c)$.

Let $a,b,c \in \mathbb{Z}$, prove that if $\gcd(a,b)=1$, then $\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)$.

algebra precalculus - Imaginary $cos^{-1}$ value significance?

When I was bored in AP Psych last year, I jokingly asked myself if there was a cosine inverse of $2$. Curious about it, I tried calculating it as follows:
$$
\begin{align*}
\cos (x) &= 2 \\

\sin (x) &= \sqrt{1 - \cos^2(x)} = \sqrt{1 - 4} = \pm i \sqrt{3}
\end{align*}
$$
Then, by Euler's formula, you have
$$
\begin{align*}
e^{ix} &= \cos (x) + i \sin (x) \\
e^{ix} &= 2 \pm\sqrt{3} \\
ix &= \ln (2 \pm \sqrt{3}) \\
x &= \boxed{-i \ln (2 \pm \sqrt{3})}

\end{align*}
$$

So, there was a way to calculate the inverse cosine of numbers whose magnitude is greater than $1$ (this was verified on Wolfram Alpha). To what extent is this kind of calculation valid? Does it have any interesting applications/implications in math, or any other subjects? Thanks. :)

Edit I just realized this is very easily explained by $2\cos (x) = e^{ix} + e^{-ix}$, but I'm still curious if this has any significance/intuition.