evaluating limits

Hi

I know I need to factor these but I just can't seem to figure out how

First


lim (x+3)/(√(9x^2-5x))
x-> infinity

Second

lim ((√(x^2+1))-x)
x ->infinity


Thanks in advance
from this calculus beginner

Quote:

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Originally Posted by calcbeg

Hi

I know I need to factor these but I just can't seem to figure out how

First


lim (x+3)/(√(9x^2-5x))
x-> infinity

Second

lim ((√(x^2+1))-x)
x ->infinity


Thanks in advance
from this calculus beginner

---

lim (x+3)/(√(9x^2-5x))
x-> infinity

This limit can be reduced to this


lim (x+3)/(|x|√(9-5/x))
x-> infinity

All you have to remember is that √x^2 = |x|

So if you factor out x^2 from under the square root you get

lim (x+3)/(√(x^2(9-5/x))
x-> infinity

Which is the same thing as

lim (x+3)/(|x|√(9-5/x))
x-> infinity

x --> infinity so x is always +


Divide everything by x
lim (1+3/x )/(√(9-5/x)) =1/3
x-> infinity

2)

lim ((√(x^2+1))-x)
x ->infinity

divide everything by x like the first question

lim ((√(1+1/x^2))-1)/(1/x)
x ->infinity

use a substitution t = 1/x

lim ((√(1+t^2)-1)/(t)
t->0

You can see that this is really the definition of a derivative just with t not h.

lim [f(x+h) - f(h)]/ h
h --> 0

f(x) = √(1+t^2)

f '(x) = 2t /2(√(1+t^2)

f ' (0) = 0 /2 = 0

So the limit = 0

Quote:

---

Originally Posted by calcbeg

Hi

I know I need to factor these but I just can't seem to figure out how

First


lim (x+3)/(√(9x^2-5x))
x-> infinity

Second

lim ((√(x^2+1))-x)
x ->infinity


Thanks in advance
from this calculus beginner

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1) If x>5/9..... .


2) If now we multiply by ,we get:


and if.......x>0.....