Thread: 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.
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[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
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Find c and n

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

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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

Q2
If then:

Originally Posted by Siron

Q2
If then:

I c, I multiplied the h to 10h that's y i couldn't get the answer.

Thanks

You're welcome!

For question 3, you notice you get the indeterminate form and the limit is rational so ...?

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

Originally Posted by Siron

You're welcome!

For question 3, you notice you get the indeterminate form 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?

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?

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