Understanding Limits

[Calc12]

Invent a function to satisfy:

a) lim x ->0+
f(x) = 3

b) lim x->0-
f(x) = -2


I thought of:

a) f(x) = x + 3

b) f(x) = x - 2


But apparently they are not good enough according to my marker.
So please, what else could it be?

Much appreciated for all help.

 

[galactus]

Invent a function to satisfy:

a) lim x ->0+
f(x) = 3

There are an endless number one could come up with.

Try this one:

\(\displaystyle \lim_{x\to 0^{+}}\frac{ln(1+3x)}{x}\)

 

[JeffM]

Invent a function to satisfy:

a) lim x ->0+
f(x) = 3

b) lim x->0-
f(x) = -2


I thought of:

a) f(x) = x + 3

b) f(x) = x - 2


But apparently they are not good enough according to my marker.
So please, what else could it be?

Much appreciated for all help.

I do not know if this will help or not, but I point out that f(x) = (x + 3) and f(x) = (x - 2) are two different functions if the domain is implied to be all real numbers. The question with its "a" seems to require an answer specifying one function.

 

[Calc12]

Oh I got confused with the wording of the problem.

I believe it's supposed to be:

Invent ONE function that satisfies both
lim x ->0+
f(x) = 3

lim x->0-
f(x) = -2

 

[JeffM]

Well then , in my humble opinion, you merely did not formulate the answer properly although you were on the right track.

f(x) such that x = (- 2) if x < 0 and x = 3 if x >= 0 is ONE such function. Does this look like what your marker was getting at?

 

[Calc12]

mhmm.. Im just having difficulties figuring it out.

I need one function that equals 3 and -2 when I plug in 0?

 

[lookagain]

Oh I got confused with the wording of the problem.

I believe it's supposed to be:

Invent ONE function that satisfies both
lim x ->0+
f(x) = 3

lim x->0-
f(x) = -2

Calc12,

what about this for the one function that covers both requirements?

\(\displaystyle f(x) = 0.5 + \frac{2.5|x|}{x}\)

 

[JeffM]

mhmm.. Im just having difficulties figuring it out.

I need one function that equals 3 and -2 when I plug in 0?

That conflicts with the definition of function. f(0) = 3 and f(0) = - 2 is an impossible contradiction.

If a is the limit of f(x) as x approaches b, f(b) is irrelevant. "Approaches" excludes "equals."

It may be that f(b) does not even exist. (Note that lookagain's function does not exist at x = 0.) If a is the limit of f(x) as x approaches b and f(b) does exist, a may not equal f(b).

The function that I specified is not continuous at b, but it is very similar to what you had come up with on your own.

The right and left limits of f(x) as x approaches b are concerned with what happens arbitrarily close to b without regard to what happens at b itself. In terms of limits, it is essential to disregard f(b).

You need to understand the difference between limit, continuity, and differentiablity. The concepts are related but different.

 

[Calc12]

thanks for all the help guys, really appreciate it.