Question

limits help

Answer 1

what do you have?

Answer 2

\[\lim_{x \rightarrow 0}x^2\] i got 0 but that messes up my problem

Answer 3

If \[\lim_{x \rightarrow 0} f(x)/x^2=5 \] find the following limits

Answer 4

using limit laws

Answer 5

see if this helps...

Answer 6

(a) \[\lim_{x \rightarrow 0}f(x)\]
(b) \[\lim_{x \rightarrow 0}f(x)/x\]

Answer 7

its number 4 of properties. so what do i do if it equals zero

Answer 8

have you done l'hospital's rule yet?

Answer 9

no i started calculus last week

Answer 10

what if f(x) = x^2
What would the limit look like then?

Answer 11

idk

Answer 12

limit as x approaches 0 of f(x)/x^2=5 if f(x)=x^2 then it wouldn't be 5

Answer 13

don't evaluate the limit. put in x^2 for for f(x) and look at it for a minute.

Answer 14

i'm lost

Answer 15

\[\lim_{x \rightarrow a}\frac{ f(x) }{ x^{2} } = 5 \Rightarrow \lim_{x \rightarrow a}\frac{5 g(x) }{ x^{2} } \text{ where } f(x) = 5g(x) \]
then
\[\lim_{x \rightarrow a}\frac{5g(x) }{ x^{2} }= 5\lim_{x \rightarrow a}\frac{ g(x) }{ x^{2} } \Rightarrow \lim_{x \rightarrow a}\frac{ g(x) }{ x^{2} }=1\]

Answer 16

a = 0
sorry

Answer 17

Nice example pgpilot. muzzammil.raza, take what I said, and combine that with pgpilot's example, which illustrates what I'm trying to get you to do

Answer 18

i am really confused right now. what i am trying to get is \[\lim_{x \rightarrow 0}f(x)\] where are you guys going

Answer 19

The problem you have is, you are trying to evaluate the limit right away. However, you can simplify the expression in the limit *before* evaluating it. limit of x/x simplifies to limit of 1.

Answer 20

\[\text{Since }\lim_{x \rightarrow a}\frac{ f(x) }{ x^{2} }=5 \Rightarrow \frac{ f(x) }{ x^{2} }=p(x)+5 \text{ where } p(x) = a_{1}x+a_{2}x^{2} +\cdots + a_{k}x^{k}\,\,k \ge 1\]
therefore
\[f(x) = p(x)\cdot x^{2} +5\cdot x^{2}\]

Answer 21

\[ k \ge 1\]

Answer 22

\[\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} p(x)\cdot x^{2} + 5\cdot x^{2}= 0 \]

Answer 23

how does this help me solve my problem

Answer 24

it's difficult getting around the division by zero... you really can't use property 4 because of this.

Answer 25

@amistre64 got anything to get around all of this?

a) should be 0 as should b)... but it's getting it all squared away.

I'm not sure I'm awake just yet.

Answer 26

\[\lim_{x \to 0} \frac{f(x)}{x^2}=5\]
im thinking that f(x) has to have a x^2 factor in it; like f(x) = 5x^2

but im still thinking thru it

Answer 27

yes... and the remainder of the thing must go to 0 when x goes to 0.

Answer 28

but not necessarily a factor, but a term

Answer 29

i kind of have to go. sorry

Answer 30

alright... sorry I couldn't help more

Answer 31

x^2 to 2x to 2 if we consider the lhop

so f'' could possibly be p(x) + 5

Answer 32

well, i mean +10 of course lol

Answer 33

you guys can still do whatever you do to solve the problem if you want. i will be back later cause i have more questions.

Answer 34

f(x) = p(x) * x^2 + 5 * x^2 where lim p(x) goes to 0 as x goes to zero

Answer 35

if it says to find the limits of a and b based upon the property that f(x)/x^2 goes to 5 as x to zero

im not sure why a or b would have to go to zero

Answer 36

suppose f(x) goes to 5... what is lim f(x)/x^2 = ?

Answer 37

is it: \[\lim \frac{f(x)}{x^2}=5\]
or: \[\frac{\lim~f(x)}{x^2}=5\]

Answer 38

lim f(x) = 5 lim x^2

lim f(x) = lim 5x^2

lim f(x)/(5x^2) = 1

Answer 39

i cant say that im real sure how to approach this ...

Answer 40

\[\text{Let } f(x) = 5\cdot g(x) \text{, then } \lim_{x \rightarrow 0}\frac{ f(x) }{ x^{2} }=\lim_{x \rightarrow 0}\frac{ 5\cdot g(x) }{ x^{2} }=5\cdot \lim_{x \rightarrow 0}\frac{ g(x) }{ x^{2} }= 5\]
\[\text{Thus } \lim_{x \rightarrow 0}\frac{ g(x) }{ x^{2} } = 1\]

Answer 41

let f(x) = 5x^2 + c1 x^3 + c2 x^4 + ... + cn x^n

in order for lHop to pan out for me, we have to have a poly of this nature on top ....

Answer 42

if we include a constant, then it doesnt zero out for the Lhop

if we include an x, then it confounds the two sided limit

Answer 43

couldn't f(x) = x^2(p(x) +5) where lim p(x) = 0 as x -> 0?

Answer 44

if we divide, we'd get p(x) + 5 and we'd need the lim of p(x) = 0.

Answer 45

it seems plausible to me

Answer 46

i beleive thats an equivalent setup yes

Answer 47

and as such, a and b would go to zero

Answer 48

Woot woot

Answer 49

what does f"(x) mean

Answer 50

Never mind it.

Answer 51

We know \(f''(x)=5\)

Answer 52

ok

Answer 53

Well, as was said before, it's possible \(f(x)=5x^2\).
It could even be \(f(x)=5x\sin(x)\)

Answer 54

In either case, I think \(f(x)\to 0\) as \(x\to 0\)

Answer 55

so...

Answer 56

There are cases where (a) is 0
And there are cases where (b) is 0

Answer 57

But I'm not sure how to prove that it is always the case. That is all I'm saying.

Answer 58

ok

Answer 59

\(\color{blue}{\text{Originally Posted by}}\) @amistre64
is it: \[\lim \frac{f(x)}{x^2}=5\]
or: \[\frac{\lim~f(x)}{x^2}=5\]
\(\color{blue}{\text{End of Quote}}\)
The second case actually isn't possible.

Answer 60

the first

Answer 61

@wio can you give one example when a isn't 0 and b isn't 0?