Determine whether each sequence converges or diverges If it converges, give the limit.
1) an=n-3/5n+2

Question

Determine whether each sequence converges or diverges If it converges, give the limit. 
1) an=n-3/5n+2

anonymous · Answer

the horizontal asymptote of 
$$f(x)=\frac{x-3}{5x+2}$$ is 
$$y=\frac{1}{5}$$

anonymous · Answer

it is the same question, only stated differently
and in a later class probably, you did this in a precalc class i am sure
it is the same thing exactly

anonymous · Answer

assuming of course that it was 
$$\lim_{n\to \infty}\frac{n-3}{5n+2}$$

kkbrookly · Answer

I am doing this in precal, but this is the first day I am working on it and I have absolutely no clue what is going on. This is what the questions look like.

anonymous · Answer

since the degree of the numerator and denominator are the same (both degree 1) the limit is the ratio of the leading coefficients

kkbrookly · Answer

Okay, so how would I determine if it is converging or diverging?

kkbrookly · Answer

Ohhhh

anonymous · Answer

if you get a number, then it converges 
since the ration of the leading coefficients is $\frac{1}{5}$ that is your limit

kkbrookly · Answer

Okay, I understand now. What about the other questions that are not in fraction form?

anonymous · Answer

second one the degree of the top is smaller than the degree of the bottom
limit is 0

kkbrookly · Answer

Why is it 0 when it is basically the same as the first equation?

anonymous · Answer

because the degree of the top is smaller than the degree of the bottom
replace n by 100 and see what you get

kkbrookly · Answer

I got .000387..

kkbrookly · Answer

So that means that the second one isn't converging?

anonymous · Answer

it does converge
it converges to 0
numbers get closer and closer to 0

kkbrookly · Answer

Alright.. I am starting to understand more. So, for something like number 3 how would I go about finding the limit? Just by plugging in a number to n?

anonymous · Answer

for number 3, the bigger n gets, the bigger $\sqrt{25+n}$ gets 
no limit ( or $\infty$)

kkbrookly · Answer

But how would I know that the bigger n gets the bigger the squareroot of 25+n gets?

anonymous · Answer

not sure what to say other than it is obvious
$\sqrt{100}=10, \sqrt{1,000,000}=1,000$ etc 
the bigger the number the bigger the square root of the number

kkbrookly · Answer

I completely understand now. Thank you for being so patient with me!

anonymous · Answer

yw
hope it is clear that the next one 
$$\sqrt{25+\frac{1}{n}}$$ has limit $5$

kkbrookly · Answer

Yes, thank you!

anonymous · Answer

yw

anonymous · Answer

eat mah poop