Convergence of Sequence with Increasing Values of n?

[karush]

View attachment 5700
ok I noticed that \(\displaystyle n \ge 0\) so we have all positive numbers and with increasing values of \(\displaystyle n\) this will go converge to zero

I can only show this by looking at a graph of the the expression. apparently the expression would have to rewritten to take the limit?

 

[Euge]

Hint: Rationalize the numerator.

 

[karush]

the expression isn't a fraction??

 

[Euge]

Any number $x$ can be written as the fraction $\frac{x}{1}$.

 

[karush]

ok will finish later got to catch a city bus;)

 

[karush]

Was the edit post option taken out?

So
$$\displaystyle
\frac{\sqrt{n+47}-\sqrt{n}}{1}
\cdot\frac{\sqrt{n+47}+\sqrt{n}}{\sqrt{n+47}+\sqrt{n}}
=\frac{47}{\sqrt{n+47}+\sqrt{n}} \\
\lim_{{n}\to{\infty}}\frac{47}{\sqrt{n+47}+\sqrt{n}}=0 $$

 

[Euge]

I see the edit post option here, but in any case, you have nothing to edit since you have no typos and your solution is correct. :D

 

[karush]

I wanted to remove post 3 and 5