Calculus Limit Help

Have a look at this Youtube video:  https://www.youtube.com/watch?v=bYZWW0fSLJU 

for a problem like this( limit to infinity) plug in a random nember and determine whether its a positive or negative. the awnser for this problem is positve infinity since (((9(1)^2)+(1))^1/2)-3(1) is positive

Find the following limit:
lim_(x->∞) (sqrt(9 x^2+x)-3 x)

sqrt(9 x^2+x)-3 x = (3 x+sqrt(9 x^2+x))/(3 x+sqrt(9 x^2+x)) (sqrt(9 x^2+x)-3 x) = x/(3 x+sqrt(9 x^2+x)):
lim_(x->∞) x/(3 x+sqrt(9 x^2+x))

Write x/(3 x+sqrt(9 x^2+x)) as 1/(3+sqrt(9 x^2+x)/x):
lim_(x->∞) 1/(3+sqrt(9 x^2+x)/x)

Using the reciprocal rule, write lim_(x->∞) 1/(3+sqrt(9 x^2+x)/x) as 1/(lim_(x->∞) (3+sqrt(9 x^2+x)/x)):
1/(lim_(x->∞) (3+sqrt(9 x^2+x)/x))

lim_(x->∞) (3+sqrt(9 x^2+x)/x) = lim_(x->∞) 3+lim_(x->∞) sqrt(9 x^2+x)/x:
1/(lim_(x->∞) 3+lim_(x->∞) sqrt(9 x^2+x)/x)

Since 3 is constant, lim_(x->∞) 3 = 3:
1/(3+lim_(x->∞) sqrt(9 x^2+x)/x)

Simplify radicals, sqrt(9 x^2+x)/x = sqrt((9 x^2+x)/x^2):
1/(3+lim_(x->∞) sqrt((9 x^2+x)/x^2))

Using the power rule, write lim_(x->∞) sqrt((9 x^2+x)/x^2) as sqrt(lim_(x->∞) (9 x^2+x)/x^2):
1/(3+sqrt(lim_(x->∞) (9 x^2+x)/x^2))

The leading term in the denominator of (9 x^2+x)/x^2 is x^2. Divide the numerator and denominator by this:
1/(3+sqrt(lim_(x->∞) (9+1/x)/1))

The expression 1/x tends to zero as x approaches ∞:
1/(3+sqrt(9))

1/(3+sqrt(9)) = 1/6:
Answer: |1/6

Thanks Guest

I learned from your answer   :)

I will just rewrite your answer in LaTex

\(\displaystyle\lim_{x\rightarrow \infty} \;\sqrt{9x^2+x}-3x\\ =\displaystyle\lim_{x\rightarrow \infty} \;\frac{\sqrt{9x^2+x}-3x}{1}\times\frac{\sqrt{9x^2+x}+3x}{\sqrt{9x^2+x}+3x}\\ =\displaystyle\lim_{x\rightarrow \infty} \;\frac{(9x^2+x)-9x^2}{\sqrt{9x^2+x}+3x}\\ =\displaystyle\lim_{x\rightarrow \infty} \;\frac{x}{\sqrt{9x^2+x}+3x}\\ =\displaystyle\lim_{x\rightarrow \infty} \;\frac{x\div x}{(\sqrt{9x^2+x}+3x)\div x}\\ =\displaystyle\lim_{x\rightarrow \infty} \;\frac{1}{\sqrt{\frac{9x^2+x}{x^2}}+3}\\ =\displaystyle\lim_{x\rightarrow \infty} \;\frac{1}{\sqrt{9+\frac{1}{x}}+3}\\ =\frac{1}{\sqrt{9+0}+3}\\ =3+3\\ =6 \)

Thank you so much. It makes sense now

Here's an alternative approach:

.

Alan,

How did you go from the second line to the third line?

Small correction:  in my fourth expression above the term beginning 1/8 should be preceded by a minus sign, not a plus sign.

Thanks Alan,

Both Chris and I have seen you do that before, I remember because we talked about it, but niether of us understood it.

I've never done binomial expansions with fractional indices....not that I remember anyway......