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