# Thread: Need help with limit

1. ## Need help with limit

Hello, I need help with few questions.

[Solved]
Lim x^2/absolute value of x =0/0
x->0

I don't know how to deal with the absolute value and factoring x^2

Its already in its lowest form therefore i substitute in zero and got the answer.
=========================
[Solved]
If f(x) = -25/2x+3
Find Lim = [f(1+h)-f(1)]/h =0/0
h->0

This is what i had done for Q2:

I substitute in f(x)

= { {-25/[2(1+h)+3]} + [25/(2(1)+3] }/ h
= { [-25/(2h+5)] + (5) } / h
= [ (-25+10h+25)/(2h+5) ] / h
= [ 10h/2h+5 ]/h
= help
======================
Find c and n

lim [(4)(x^9-11)(x^3+8)] / [(c)(x^n-8)(x^2+17)] = 4/4 = 1
x-> infinity

=======================

let f(x) = [(3)(x^9-12)(x^3+3)] and let g(x) = [(c)(x^n-9)(x^2+16)] with c cannot = 0
Lim
x-> infinity f(x)/g(x)= infinity

Implies that n (greater or equal or less than) (?)

for the last two question i have no idea how to start.
Thank You

2. ## Re: Need help with limit

Q2
If $f(x)=\frac{-25}{2x+3}$ then:
$\lim_{h\to 0} \frac{\frac{-25}{2(1+h)+3}+5}{h}$
$=\lim_{h \to 0} \frac{\frac{-25+5(5+2h)}{5+2h}}{h}$
$=\lim_{h\to 0} \frac{\frac{10h}{5+2h}}{h}$
$=\lim_{h\to 0} \frac{10h}{(5+2h)h}$
$=\lim_{h \to 0} \frac{10}{5+2h}=2$

3. ## Re: Need help with limit

Originally Posted by Siron
Q2
If $f(x)=\frac{-25}{2x+3}$ then:
$\lim_{h\to 0} \frac{\frac{-25}{2(1+h)+3}+5}{h}$
$=\lim_{h \to 0} \frac{\frac{-25+5(5+2h)}{5+2h}}{h}$
$=\lim_{h\to 0} \frac{\frac{10h}{5+2h}}{h}$
$=\lim_{h\to 0} \frac{10h}{(5+2h)h}$
$=\lim_{h \to 0} \frac{10}{5+2h}=2$
I c, I multiplied the h to 10h that's y i couldn't get the answer.

Thanks

4. ## Re: Need help with limit

You're welcome!

For question 3, you notice you get the indeterminate form $\frac{\infty}{\infty}$ and the limit is rational so ...?

5. ## Re: Need help with limit

As far as the first one is concerned, are you aware that |x|= x if $x\ge 0$ and |x|= -x if x< 0? Use those two cases to determine what |x|/x is and what the limit is.

6. ## Re: Need help with limit

Originally Posted by Siron
You're welcome!

For question 3, you notice you get the indeterminate form $\frac{\infty}{\infty}$ and the limit is rational so ...?
For indeterminate form of infinity/infinity of a rational function you divide the numerator and denominator by x^n where n is the highest power present in the denominator.

Both Q3 and Q4 have x^n in the denominator. There isn't a specific value. I cannot just divide by x^n can I? The question gave the limit. How do i work backwards?

7. ## Re: Need help with limit

Yes, so i do both cases. Therefore (0^2)/(0) = 0/0 = 0
(0^2)/-(0) = 0/0 = 0
The Lim is zero.

If the x is not approaching zero, can there be more than one limit?

8. ## Re: Need help with limit

I managed to solved the rest myself, thank you all.