limits - Evaluate $lim_{x to -infty} left(frac{sqrt{1+x^2}-x}{x} right)$

Evaluate $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$

I tried by taking $x^2$ out of the root by taking it common.

i.e: $$\lim_{x \to -\infty} \left(\frac{x\sqrt{\frac{1}{x^2}+1}-x}{x} \right)$$
and then cancelling the x in numerator and denominator

$$\lim_{x \to -\infty} \left(\frac{\sqrt{\frac{1}{x^2}+1}-1}{1} \right)$$

then substituting $x= -\infty$ in the equation, we get,
$$\lim_{x \to -\infty} \left(\frac{\sqrt{0+1}-1}{1} \right)$$
which equals to $0$. But it is not the correct answer.

What have I done wrong.

Answer

When $x\lt 0$, we have
$$\sqrt{1+x^2}\not =x\sqrt{\frac{1}{x^2}+1}.$$
(Note that LHS is positive and that RHS is negative!)

You can set $-x=t\gt 0$.
$$\lim_{x\to -\infty}\frac{\sqrt{1+x^2}-x}{x}=\lim_{t\to\infty}\frac{\sqrt{1+(-t)^2}+t}{-t}=\lim_{t\to\infty}\left(-\sqrt{\frac{1+t^2}{t^2}}-1\right)=-2.$$

calculus - How can I calculate the limit of exponential divided by factorial?

I suspect this limit is 0, but how can I prove it?

$$\lim_{n \to +\infty} \frac{2^{n}}{n!}$$

Answer

The easiest way to do this is to do the following: Assume $n \ge 4$. Then $$0 \le \frac{2^n}{n!} = \prod_{i=1}^n \frac{2}{i} = \frac{2\cdot 2\cdot 2}{1 \cdot 2 \cdot 3} \cdot \prod_{i=4}^n \frac{2}{i} \le \frac{8}{6} \cdot \prod_{i=1}^n \frac{2}{4} = \frac{8}{6 \cdot 2^{n-3}}.$$ Applying the squeeze theorem gives the result.

algebra precalculus - Solve ......................................................................

This is what I did

Can anyone tell me what's wrong me or the question?

Answer

You did the case $y=4x$, and got answers $x=1$ and $x=-\frac{9}{4}$. Neither is a solution of the original equation. It is easy to verify that when $y=4x$ we have $\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}=-\frac{3}{2}$.

When you squared, you introduced extraneous roots. Only $x=4y$ leads to a solution of the system.

limits - Evaluate $lim_{n to infty }frac{(n!)^{1/n}}{n}$.

Evaluate
$$\lim_{n \to \infty }\frac{(n!)^{1/n}}{n}.$$

Can anyone help me with this? I have no idea how to start with. Thank you.

Answer

Let's work it out elementarily by wisely applying Cauchy-d'Alembert criterion:

$$\lim_{n\to\infty} \frac{n!^{\frac{1}{n}}}{n}=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty} \frac{(n+1)!}{(n+1)^{(n+1)}}\cdot \frac{n^{n}}{n!} = \lim_{n\to\infty} \frac{n^{n}}{(n+1)^{n}} =\lim_{n\to\infty} \frac{1}{\left(1+\frac{1}{n}\right)^{n}}=\frac{1}{e}. $$

Also notice that by applying Stolz–Cesàro theorem you get the celebre limit:

$$\lim_{n\to\infty} (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}} = \frac{1}{e}.$$

The sequence $L_{n} = (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}}$ is called Lalescu sequence, after the name of a great Romanian mathematician, Traian Lalescu.

Q.E.D.

elementary number theory - Prove that $gcd(a,b) = gcd (a+b, gcd(a,b))$

I started by saying that $\gcd(a,b) = d_1$ and $\gcd(a+b,\gcd(a,b)) = d_2$

Then I tried to show that $\ d_1 \ge d_2, d_1 \le d_2$.

I know that $\ d_2 | \gcd(a+b, d_1)$ hence $\ d_2 \le d_1 $.

How do I prove that $\ d_2 \ge d_1$ ?

Answer

If $\gcd(a,b)=d_1$ then $a = d_1 x$ and $b= d_1 y$, where $x,y$ are integers. Consequently,
$$\gcd(a+b,\gcd(a,b))=\gcd(d_1(x+y),d_1) = d_1\gcd(x+y,1)=d_1.$$

A proof by induction and trigonometry

Do you know how to prove that $\displaystyle\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ using induction?

---

I have tried with $n = 1$ which gives $\cos \frac{x}{2} = \frac{\sin(nx)}{(2\sin1/2x)}$

I am not sure on how to expand with the trigonometric formulas.

With $n= p+1$ I get LHS: $\cos(2(n+1)-1)$ which I summaries to $\cos(2n+1)$ which should be $\cos 2n \cos 1-\sin 2n \sin1$ plus the RHS $\frac{\sin(nx)}{(2\sin 1/2x)}$

RHS $p+ 1 = \frac{\sin(n+1x)}{(2\sin 1/2x)}$

Any ideas on how to proceed would be very helpful.

Answer

If $$\sum_{r=1}^n\cos\dfrac{(2r-1)}2x=\dfrac{\sin nx}{2\sin\dfrac x2}$$ holds true for $n=m$

$$\sum_{r=1}^{m+1}\cos\dfrac{(2r-1)}2=\dfrac{\sin mx}{2\sin\dfrac x2}+\cos\dfrac{[2(m+1)-1]}2x$$

$$=\dfrac{\sin mx+2\sin\dfrac x2\cos\dfrac{(2m+1)}2x}{2\sin\dfrac x2}$$

Using Werner's formula, $2\sin\dfrac x2\cos\dfrac{(2m+1)}2x=\sin(m+1)x-\sin mx$

real analysis - How to construct a bijection from $(0, 1)$ to $[0, 1]$?

I wonder if I can cut the interval $(0,1)$ into three pieces:
$(0, \frac{1}{3})\cup(\frac{1}{3},\frac{2}{3})\cup(\frac{2}{3},1)$, in which I'm able to map point $\frac{1}{3}$ and $\frac{2}{3}$ to $0$ and $1$ respectively.

Now the question remained is how to build a bijection mapping from those three intervels to $(0,1)$.

Or, my method just goes in a wrong direction. Any correct approaches?

Answer

Consider the sequence
$$
\frac 12, \frac 13, \frac 14, \frac 15, \dots, \frac 1n, \dots
$$
Map every other point $f : (0,1) \to [0,1]$ that is not in this sequence to itself, and then map the above sequence to the corresponding points in this one :
$$

0, 1, \frac 12, \frac 13, \frac 14, \dots.
$$
In other words, map $\frac 12$ to $0$, $\frac 13$ to $1$, and then map $\frac 1n$ to $\frac 1{n-2}$ for $n \ge 4$.

The reason why you can map some set into some bigger set bijectively is precisely because they are infinite, so you must exploit this fact. If you don't, you have no chance.

Now to answer your actual question, the trick you try to use doesn't feel relevant to me ; I'm not saying there is absolutely no way it could work, because I actually know there is, since your set (the union of the three intervals) and the interval $(0,1)$ have the same cardinality. The problem with your idea is that I don't think a construction will naturally come out of it. In general, to map bijectively a set into a bigger one you must be "moving things around", so I expect any fairly understandable construction involving your idea to be similar to the one I've shown you.

Hope that helps,