Prove that the limit constant times a function

[gothloli]

Homework Statement

the question is: Prove that if lim$_{x→0}$ f(x)/x = l and b≠0, then lim$_{x→0}$ f(bx)/x = bl

Homework Equations

Hint: write f(bx)/x = b[f(bx)/bx]
properties of limits, delta epsiolon...

The Attempt at a Solution

I assume that I can use the property of limits that is lim(x->a) (f*g)(x) = l*m
I can make b one function and f(x)/x another, and hence use the above property. But I honestly don't think that's right, don't know where to go from here

 

[SammyS]

Homework Statement

the question is: Prove that if lim$_{x→0}$ f(x)/x = l and b≠0, then lim$_{x→0}$ f(bx)/x = bl

Homework Equations

Hint: write f(bx)/x = b[f(bx)/bx]
properties of limits, delta epsiolon...

The Attempt at a Solution

I assume that I can use the property of limits that is lim(x->a) (f*g)(x) = l*m
I can make b one function and f(x)/x another, and hence use the above property. But I honestly don't think that's right, don't know where to go from here

Where to go from here? ... good question.

Do you suppose you are to do an ε - δ proof, or can you do something with substitution?

 

[raopeng]

I think that would suffice, after all x is an infinitesimal and multiplied by a constant won't change anything. Just like infinitesimal is a number smaller than any given number ε, so it is perfectly okey to have an infinitesimal smaller than $\frac{ε}{b}$ I think...

 

[gothloli]

Where to go from here? ... good question.

Do you suppose you are to do an ε - δ proof, or can you do something with substitution?

You could use ε - δ proof, but can also use the properties of limits to prove this, like sum of functions is equal to sum of their limits, and product and quotient properties, at least that's what my book says. I don't get the hint given in my book though.

I think that would suffice, after all x is an infinitesimal and multiplied by a constant won't change anything. Just like infinitesimal is a number smaller than any given number ε, so it is perfectly okey to have an infinitesimal smaller than $\frac{ε}{b}$ I think...

can you explain that further, I'm kind of new to all this, only in first year.

 

[HallsofIvy]

Let u= bx so that x= u/b. Then
$$\frac{f(bx)}{x}= \frac{f(u)}{\frac{u}{b}}= \frac{bf(u)}{u}= b\frac{f(u)}{u}$$
and, of course, as x goes to 0, so does u.

 

[gothloli]

Let u= bx so that x= u/b. Then
$$\frac{f(bx)}{x}= \frac{f(u)}{\frac{u}{b}}= \frac{bf(u)}{u}= b\frac{f(u)}{u}$$
and, of course, as x goes to 0, so does u.

I don't get it also what happens if b=0

 

[raopeng]

If I remember it correctly, an infinitesimal is a quantity that is smaller than any number given, hence here $lim_{x→0} x$ is smaller than an arbitrary |ε|. where ε can be as small as 0.000000000000001 or -0.000000000000000000000001. So as we can see when x is indefinitely small, if multiplied by a constant b, it is still very very small(the quantity won't change much if you multiply 0.0000000000000000001 by 6 for example), so we can redefine $lim_{x→0} x$ as smaller than $|\frac{ε}{b}|$(ε is still any arbitrary value and clearly it is equivalent with the above definition). Or in a less dogmatic sense, we can say $lim_{x→0} x$ is very small(can be regarded as 0) so multiply it with any constant it is still almost zero(0 multiplies with any number still gives 0), and then we can use the substitution to obtain the results as HallsofIvy said.

If b = 0, then for any values of x, f(bx) degrades into f(0).

 

[raopeng]

for example, $lim_{x→0}\frac{sin bx}{x} = b$ for b is not zero. If b is 0, then the whole expression becomes $lim_{x→0}\frac{sin 0}{x} = 0$(because anything multiplies zero only gives zero). You can do a Taylor expansion to verify this I think.

 

[gothloli]

Let u= bx so that x= u/b. Then
$$\frac{f(bx)}{x}= \frac{f(u)}{\frac{u}{b}}= \frac{bf(u)}{u}= b\frac{f(u)}{u}$$
and, of course, as x goes to 0, so does u.

Okay now I get it, after contemplating for a long time, lol thanks.

If I remember it correctly, an infinitesimal is a quantity that is smaller than any number given, hence here $lim_{x→0} x$ is smaller than an arbitrary |ε|. where ε can be as small as 0.000000000000001 or -0.000000000000000000000001. So as we can see when x is indefinitely small, if multiplied by a constant b, it is still very very small(the quantity won't change much if you multiply 0.0000000000000000001 by 6 for example), so we can redefine $lim_{x→0} x$ as smaller than $|\frac{ε}{b}|$(ε is still any arbitrary value and clearly it is equivalent with the above definition). Or in a less dogmatic sense, we can say $lim_{x→0} x$ is very small(can be regarded as 0) so multiply it with any constant it is still almost zero(0 multiplies with any number still gives 0), and then we can use the substitution to obtain the results as HallsofIvy said.

If b = 0, then for any values of x, f(bx) degrades into f(0).

Yes and thanks for the help!

 

[Mark44]

I don't get it also what happens if b=0

In your first post, you say that b ≠ 0.

Famous saying:

If all else fails, read the instructions.