Question

Limits help..

Answer 1

\[\Huge \lim_{x \rightarrow \infty} \frac{3x^m+4x^{m-2}+3}{4x^n+3x^{n-2}+5}\]

Answer 2

A) 0 if m>n
B) 0 if m<n
C) infinity if m>n
D) infinity if m<n

Answer 3

Proceeding stepwise,
\[\Huge \lim_{x \rightarrow \infty} \frac{3x^m+\frac{1}{4}x^{m}+3}{4x^n+\frac{1}{9}x^{n}+5}\]

Answer 4

We generally factor out to the highest coefficient,
but not quite sure

Answer 5

Sorry! One question, why is \(\frac{1}{9}x^n = 3x^{n-2}\)?

Answer 6

that should be 1/3 my bad

Answer 7

\[\Huge \lim_{x \rightarrow \infty} \frac{3x^m+\frac{1}{4}x^{m}+3}{4x^n+\frac{1}{3}x^{n}+5}\]

Answer 8

@terenzreignz @oldrin.bataku @experimentX

Answer 9

Still don't understand :| Isn't it
\[3x^{n-2} = \frac{3x^n}{x \times x}=\frac{3x^n}{x^2}\]?

Answer 10

oh lol,hmm..yeah :P

Answer 11

\[\Huge \lim_{x \rightarrow \infty} \frac{3x^m+\frac{4x^{m}}{x^2}+3}{4x^n+\frac{3x^{n}}{x^2}+5}\]

Answer 12

\[ \lim_{x \rightarrow \infty} \frac{3x^m+4x^{m-2}+3}{4x^n+3x^{n-2}+5}\]\[=\large \lim_{x \rightarrow \infty} \frac{\frac{3x^m+4x^{m-2}+3}{x^n}}{\frac{4x^n+3x^{n-2}+5}{x^n}}\]If n>m, \(\frac{3x^m}{x^n} \rightarrow 0\) when \(x\rightarrow \infty\)

Answer 13

yeah that will tend to 0

Answer 14

So the whole thing -> 0?!

Answer 15

possibly?

Answer 16

O_O

Answer 17

\[=\large \lim_{x \rightarrow \infty} \frac{\frac{3x^m+4x^{m-2}+3}{x^n}}{\frac{4x^n+3x^{n-2}+5}{x^n}}\]\[=\large \lim_{x \rightarrow \infty} \frac{0+0+0}{4+0+0}\]if n>0?

Answer 18

yup,seems okay

Answer 19

If m>n
\[ \lim_{x \rightarrow \infty} \frac{3x^m+4x^{m-2}+3}{4x^n+3x^{n-2}+5}\]\[ = \large\lim_{x \rightarrow \infty} \frac{\frac{3x^m+4x^{m-2}+3}{x^n}}{\frac{4x^n+3x^{n-2}+5}{x^n}}\]
denominator -> 4
\(\frac{3x^m}{x^n} \rightarrow \infty\) when \(x \rightarrow \infty\) ?

Answer 20

seems true :)

Answer 21

I need @hartnn to confirm - he is my hope ><!

Answer 22

yeah correct, 0 when n>m

Answer 23

it's multiple choice

Answer 24

Can't be (A) because \(m>n\) means the numerator dominates the denominator in the long run

Answer 25

(B) looks correct though!

Answer 26

$$3x^m/4x^n=3/4\ x^{m-n}$$If \(m>n\) we have \(\infty\) yet if \(m<n\) we have \(0\)!

Answer 27

B and C both look correct to me...

Answer 28

My first step is this?
\[\LARGE \lim_{x \rightarrow \infty} \frac{\frac{3x^m}{x^n}+\frac{4x^{m-2}}{x^n}+\frac{3}{x^n}}{\frac{4x^n}{x^n}+\frac{3x^{n-2}}{x^n}+\frac{5}{x^n}}\]

Answer 29

now.if n>m then it should be 0 and if m>n then infinity?

Answer 30

(C) looks good too yeah

Answer 31

do we even need to divide it by x^n? or can we make it out simply from the question too

Answer 32

@oldrin.bataku ? @Callisto ?

Answer 33

"now.if n>m then it should be 0 and if m>n then infinity?" <- I think so :|

Answer 34

but do we need to divide by x^n? in the 1st step,that can be seen from the question too

Answer 35

Hmm.. If it's just an MC question, you can skip as many steps as you like. :|
Though, I prefer writing this step...

Answer 36

so a single step and rest is observation correct?

Answer 37

You can say so...

Answer 38

okay,thanks! :D

Answer 39

Welcome :)