Limits help

• Sep 4th 2006, 05:20 PM
Nimmy
Limits help
1. Find the lim x goes to 4 on the postive side of the function x-5/(4-x^3)(x-7)^2.
2. Find the lim x goes to postive Pi/2 of the function tan x.
3. Find the lim x goes to infinity (2x+1)/x^2-(2x-1)/x^2.
4. Find the lim x goes to infinity (x^2)/(2x+1) - (x^2)/(2x-1)
5. Fiind the lim x goes to infinity x^3/3^x
• Sep 4th 2006, 05:55 PM
ThePerfectHacker
Quote:

Originally Posted by Nimmy
1. Find the lim x goes to 4 on the postive side of the function x-5/(4-x^3)(x-7)^2.[/tex]

It is countinous, simplfy substitute 4 for x
Quote:

2. Find the lim x goes to postive Pi/2 of the function tan x.
The limits does not exist, however $+\infty$

Quote:

5. Fiind the lim x goes to infinity x^3/3^x
Exponential always overtake polynomials so the limit is zero.
• Sep 4th 2006, 08:57 PM
CaptainBlack
Quote:

Originally Posted by Nimmy
4. Find the lim x goes to infinity (x^2)/(2x+1) - (x^2)/(2x-1)

$
\frac{x^2}{2x+1} - \frac{x^2}{2x-1}=\frac{x^2(2x-1)-x^2(2x+1))}{4x^2-1}=-\frac{2x}{4x^2-1}$
$=\frac{2}{4+1/x^2}
$

Therefore:

$
\lim_{x \to \infty} \frac{x^2}{2x+1} - \frac{x^2}{2x-1}=$
$\lim_{x \to \infty}\frac{2}{4+1/x^2}=1/2
$

RonL
• Sep 6th 2006, 07:41 PM
Nimmy
Math Help Homework
Find the lim x goes to inifinity (2/3x-4)? and also for the equation of square root of 4x+1.
• Sep 6th 2006, 07:46 PM
Jameson
Is the first one $\lim_{x \rightarrow \infty} \frac{2}{3}x-4$ or $\lim_{x \rightarrow \infty}\frac{2}{3x}-4$?
• Sep 6th 2006, 07:48 PM
Nimmy
Ok nvr mind about that problem but i do need help on square root of 4x+1 using the def. of dervitaitve. Dont use the calculus way!!!
• Sep 7th 2006, 05:15 AM
topsquark
Quote:

Originally Posted by Nimmy
Ok nvr mind about that problem but i do need help on square root of 4x+1 using the def. of dervitaitve. Dont use the calculus way!!!

How can we not use Calculus if we are using the definition of a derivative??

-Dan

For the record:
$\frac{d}{dx}\sqrt{4x+1} = \frac{2}{\sqrt{4x+1}}$ which is positive for all x in its domain, so we see that $\sqrt{4x+1}$ is monotonically increasing on $[ -\frac{1}{4}, \infty )$ and it has no horizontal asymptote. Thus $\lim_{x \to \infty} \sqrt{4x+1} \to \infty$.
• Sep 7th 2006, 05:23 AM
CaptainBlack
Quote:

Originally Posted by Nimmy
Find the lim x goes to inifinity (2/3x-4)? and also for the equation of square root of 4x+1.

What about $\sqrt{4x+1}$ is it that you want help with?

RonL
• Sep 10th 2006, 05:52 PM
Nimmy
Math Help Homework
Find a and b so that the slope of the tangent line to f(x)= ax^2+bx (1,3) is -2.
• Sep 10th 2006, 06:01 PM
topsquark
Quote:

Originally Posted by Nimmy
Find a and b so that the slope of the tangent line to f(x)= ax^2+bx (1,3) is -2.

When you are asking a new question, please start a new thread. It's much less confusing that way, and much easier for someone else to find help on a similar problem.

The slope of the tangent line to $f(x) = ax^2 + bx$ is the first derivative $f'(x) = 2ax + b$. At the point (1,3) $f'(1) = 2a + b = -2$. There are many possible solutions to this equation, so I'll simply say that you can pick any "a" value and the required "b" value will then be $b = -2a - 2$.

-Dan
• Sep 10th 2006, 06:13 PM
Nimmy
So do u basically plug in -2a-2 into f'(x) or f(x)?
• Sep 10th 2006, 06:30 PM
ThePerfectHacker