3 limits that I want to check

[FrEaKmAn]

Hi,

hopefully I will get some response as this being my first post on this forum

I have 3 limits that I'm not so sure if I solved them correctly...

this first is not problem, now with second one (I hope you will understand)

\(\displaystyle (2^n * 2 + 3^n * 3)/(2^n+3^n)=(2^n+n/3^n + 3^n*3/3^n)/(2^n/3^n + 3^n/3^n)=3/1=3\)

I have no idea how to start (c), maybe something like this

\(\displaystyle (n+1)^(1/2)-n^(1/2)=n^(1/2)+n/2+1^(1/2)-n^(1/2)=n/2 + 1\) but does this count as infinity?

about (d) maybe to much to write, but possible that we convert everything to same root = 12, then remove it and divide everything with biggest number (n^18) to get result 1

 

[galactus]

The first one is easy, -1/2.


The second is correct. Good work


The third one can be found by using the conjugate. Which I think you wered doing

\(\displaystyle \L\\\lim_{n\to\infty}\frac{\sqrt{n+1}-\sqrt{n}}{1}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\)

\(\displaystyle \L\\\lim_{n\to\infty}\frac{1}{\sqrt{n+1}+\sqrt{n}}=0\)

The fourth is correct, 1. Good.

 

[FrEaKmAn]

I think I got (mathematica program) infinity for third one, but you said 0, any ideas?

 

[galactus]

For c, you can see as n gets larger, the limit approaches 0.

e.g. 1/n, as n gets larger the limit is 0.

Intuitively, take a look \(\displaystyle \L\\\sqrt{100000000001}-\sqrt{100000000000}=0.00000158\)

It gets smaller and smaller. The difference in the numbers is only 1, so as it gets bigger the difference gets nearer to 0.

I also ran it through tech and got 0.