# evaluating limits

• Feb 8th 2009, 02:59 PM
calcbeg
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

from this calculus beginner
• Feb 8th 2009, 03:05 PM
Banned for attempted hacking
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

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
• Feb 9th 2009, 02:52 PM
archidi
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

1) If x>5/9..... $\lim_{x\to\infty}{\frac{\ x+3}{\sqrt{9x^2-5x}}}=\lim_{x\to\infty}{\frac{\ 1+\frac{\ 3}{x}}{\sqrt{9-\frac{\ 5}{x}}}}=\frac{\ 1+3.0}{\sqrt{9-5.0}}=\frac{\ 1}{3}$.
2) If now we multiply $\sqrt{x^2+1} -x$ by $\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$,we get:
$\frac{\ 1}{\sqrt{x^2+1}+x}$ and if.......x>0..... $\lim_{x\to\infty}{\frac{\ 1}{\sqrt{x^2+1}+x}}=\lim_{x\to\infty}{\frac{\frac{ \ 1}{x}}{\sqrt{1+\frac{\ 1}{x^2}}+1}}=\frac{\ 0}{\sqrt{1+0}+1}=0$