Q2
If then:
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
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?