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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- Feb 8th 2009, 02:59 PMcalcbegevaluating 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 - Feb 8th 2009, 03:05 PMBanned for attempted hacking

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 PMarchidi

1) If x>5/9..... $\displaystyle \lim_{x\to\infty}{\frac{\ x+3}{\sqrt{9x^2-5x}}}=\lim_{x\to\infty}{\frac{\ 1+\frac{\ 3}{x}}{\sqrt{9-\frac{\ 5}{x}}}}=\frac{\ 1+3.0}{\sqrt{9-5.0}}=\frac{\ 1}{3}$.

2) If now we multiply $\displaystyle \sqrt{x^2+1} -x$ by $\displaystyle \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$,we get:

$\displaystyle \frac{\ 1}{\sqrt{x^2+1}+x}$ and if.......x>0..... $\displaystyle \lim_{x\to\infty}{\frac{\ 1}{\sqrt{x^2+1}+x}}=\lim_{x\to\infty}{\frac{\frac{ \ 1}{x}}{\sqrt{1+\frac{\ 1}{x^2}}+1}}=\frac{\ 0}{\sqrt{1+0}+1}=0$