# Limits of Functions?

• Oct 2nd 2009, 06:11 PM
dark-ryder341
Limits of Functions?
Hello everyone,

Well, our class recently started learning limits, and I'm having some troubles with two of the questions from the homework.

The first question is:

State the value of the limit of the function: f(x) = 4/5+e^(1/x)

lim f(x) for x --> 0+

I got 4/5+e^(1/0+) = 4/5+e^(infinity) = 4/5+infinity....I said the answer was infinity, but that was wrong.

The other question involves finding the limits of functions that have square roots in them. I was wondering if there's any 'general' way of solving those sorts of functions, since I have four of those questions, and I'm unsure of how to go about those.

Anyhow, the second question is: Find the limit for:

lim for x --> -infinity squareroot(9x^6-x)/x^3+6

I really have no idea how to go about this question, so any help is greatly appreciated!
• Oct 2nd 2009, 06:22 PM
DeMath
Quote:

Originally Posted by dark-ryder341
Hello everyone,

Well, our class recently started learning limits, and I'm having some troubles with two of the questions from the homework.

The first question is:

State the value of the limit of the function: f(x) = 4/5+e^(1/x)

lim f(x) for x --> 0+

I got 4/5+e^(1/0+) = 4/5+e^(infinity) = 4/5+infinity....I said the answer was infinity, but that was wrong.

Maybe you given this function $\displaystyle f\left( x \right) = \frac{4}{{5 + {e^{1/x}}}}$ ??
• Oct 2nd 2009, 06:27 PM
dark-ryder341
Yes, that was the function I was given. Sorry, I don't know how to write it out properly like that. I'm just not sure how to solve it...
• Oct 2nd 2009, 06:34 PM
DeMath
Quote:

Originally Posted by dark-ryder341
Yes, that was the function I was given. Sorry, I don't know how to write it out properly like that. I'm just not sure how to solve it...

Then

$\displaystyle f\left( x \right) = \frac{4}{{5 + {e^{1/x}}}}$

$\displaystyle \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{4}{{5 + {e^{1/x}}}} = \frac{4} {{5 + {e^\infty }}} = \frac{4}{\infty } = 0.$
• Oct 2nd 2009, 06:39 PM
DeMath
Quote:

Originally Posted by dark-ryder341
Hello everyone,

Well, our class recently started learning limits, and I'm having some troubles with two of the questions from the homework.

The other question involves finding the limits of functions that have square roots in them. I was wondering if there's any 'general' way of solving those sorts of functions, since I have four of those questions, and I'm unsure of how to go about those.

Anyhow, the second question is: Find the limit for:

lim for x --> -infinity squareroot(9x^6-x)/x^3+6

I really have no idea how to go about this question, so any help is greatly appreciated!

$\displaystyle \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {9{x^6} - x} }} {{{x^3} + 6}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| {{x^3}} \right|\sqrt {9 - \frac{1}{{{x^5}}}} }}{{{x^3}\left( {1 + \frac{6} {{{x^3}}}} \right)}} = - \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {9 - \frac{1}{{{x^5}}}} }}{{1 + \frac{6}{{{x^3}}}}} = - \frac{{\sqrt {9 + 0} }}{{1 - 0}} = - 3.$
• Oct 2nd 2009, 07:28 PM
dark-ryder341
Thanks very much for your help! Also, I was wondering for this question:

http://i35.tinypic.com/ohlc2h.jpg

How were the values of f(x) obtained? I tried plugging in the x values into the function, but got different numbers.
• Oct 2nd 2009, 08:42 PM
redsoxfan325
Are you sure you didn't round at all in the middle? I'm plugging in x and getting out what they have for f(x).
• Oct 2nd 2009, 09:33 PM
dark-ryder341
Nope, I didn't round...when I plugged in 1 for x, I got 4.582575695...maybe I'm entering it in my calculator wrong?

squareroot(1+25-5/1) ?
• Oct 2nd 2009, 09:37 PM
mr fantastic
Quote:

Originally Posted by dark-ryder341
Nope, I didn't round...when I plugged in 1 for x, I got 4.582575695...maybe I'm entering it in my calculator wrong?

squareroot(1+25-5/1) ?

More brackets are needed. Perhaps making correct use of them will give you the correct answer.

(Sqrt(1 + 25) - 5)/1

etc.