Question

Limit question:
If
\[\lim_{x \rightarrow 0} \frac{ f(x) }{x^2}=1\], what is \[\lim_{x \rightarrow 0} f(x)\]?

Answer 1

I'm going to go with 2.

You're going to get indeterminate forms until you do L'Hospitals down to get the constants. In this case the constant in the denominator is 2. So for the limit to be 1, then the limit of the numerator must be 2.

Answer 2

assuming f(x) limit = 0 as it approaches x=0 you get

0/0 which indeterminate

take the derivative of x^2

f'(x)=f'(x)
\[g(x)=x^2\]
\[g'(x)=2x\]
since you still get 0 on the bottom for 0 take another derivative

2=g''(x)

f''(x)=2
f'(x)=2x+c
\[f(x)=x^2+cx+c_1\]

Answer 3

limit has to be 0 otherwise you wont have an indeterminate form

Answer 4

Would you mind explaining why you keep taking derivative for denominator?

Answer 5

because if you get anything on the top for a limit you'll get
\[\frac{a}{0}\]

Answer 6

which is undefined correct

Answer 7

so the only time you can get an actual answer is if you have 0/0 because then you can use l'hopitals rule

Answer 8

So, we have to take derivative of the denominator (g(x)) so that denominator \(\ne\) 0, Right?
Here comes to another problem..
Why, when you get g''(x) = 2, f''(x) would also be 2?
I'm sorry but I haven't learnt L'Hospitals rule yet..

Answer 9

yep

Answer 10

because if at g''(x)= 2, in order to get 1, you have to have a constant 2

\[\frac{2}{2}=1\]

Answer 11

what calculus are you in?

Answer 12

I'm not from the US.

Answer 13

calculus 2 will use this method called l'hopitals, where if you get indeterminate forms, there still exists a limit , so you use derivatives to find it. If you're in calculus 1 you can simply think of it as such

if the limit as x->0 of f(x)/x^2=1, What possibly could the f(x) be to cancel out the x^2 on the bottom

if \[x^2\] is f(x) you get

\[\frac{x^2}{x^2}=1\]

the limit as x->0 of 1 is always 1

Answer 14

I'm also thinking in the second way, but I'm not quite sure how to work the answer out...
Wait.. If x^2 is not cancel, wouldn't the limit of f(x) be 0?

Answer 15

this is a shortcut for limits that have the same power polynomial in the numerator and denominator

\[\lim_ {x\rightarrow 0}\frac{2x^2}{x^2}=2\]

always will =2 no matter what

this is the same rule where as

\[\lim _{x \rightarrow 0}\frac{x^2}{x^2}=1|_{x=0}=1\]

Answer 16

if it didn't cancel you'd get

\[\lim _{x \rightarrow 0} \frac{f(x)}{x^2}|_{x=0}=\frac{f(x)}{0}=undefined\]

Answer 17

In that case, \[\lim_{x \rightarrow 0} f(x) = x^2 = 0^2 = 0\], if we use that shortcut?

Answer 18

Wait...
I can think of two possible cases for f(x)
1. f(x) = x^2
2. f(x) = (x^2) P(x), where there is a constant term in P(x)

Answer 19

yes the limit would have to be 0 because the only way that function as the limit goes to 0 equals 1 is if the top is in quadratic form which is why i posted

\[x^2+cx+c_1\]
it's the same as

\[x^2+bx+c\]

Answer 20

But how can we determine if the constant term is 0? If the constant term c = 1, limit of f(x)/x when x-> 0 is still 1, isn't it?

Answer 21

in a case of the polynomial form

Example 2:

\[\lim _{x \rightarrow 0} \frac{x^2+15x+100}{x^2}\]

this would = 1, the long way of solving this would be to take the denominator and expand and then using l'hopital

\[\lim_ {x \rightarrow 0} \frac{x^2}{x^2}+\frac{15x}{x^2}+\frac{100}{x^2}=1+\frac{15}{x}+\frac{100}{x^2}\]

Answer 22

But it is not specified if f(x) is a polynomial of degree n \(\ge\) 2..

Answer 23

What I mean was if \(f(x) = x^2 (x^n + x^{n-1} +...+c)\)
\[\lim_{x \rightarrow 0} \frac{f(x)}{x^2} = \frac{x^2 (x^n + x^{n-1} +...+c)}{x^2} = x^n + x^{n-1} +...+c = c\] Of course, c must be 1 in this case.

Answer 24

Oh well, I'm moving on the wrong track. It doesn't matter what c is when we have to find the limit of f(x). Anyway, it must be zero!

Answer 25

Thank you so much for your help!