Limit of Division of Functions

[eric_f]

I have a homework problem that I can't quite figure out:

Give an example of functions \(\displaystyle y=f(x)\) and \(\displaystyle y=g(x)\) such that \(\displaystyle \lim_{ x\rightarrow 1}f(x)=0\), but \(\displaystyle \lim_{ x\rightarrow 1} \frac{f(x)}{g(x)}=8\).

So far, the only functions that I have been able to conjure up are these:

\(\displaystyle f(x)=16x-16\)
\(\displaystyle g(x)=x^2-1\)

so

\(\displaystyle \lim_{x\rightarrow 1} \frac{f(x)}{g(x)}=\lim_{ x\rightarrow 1}\frac{16x-16}{x^2-1}=\lim_{ x\rightarrow 1}\frac{16(x-1)}{(x+1)(x-1)}=\lim_{ x\rightarrow 1}\frac{16}{x+1}=\)

\(\displaystyle \lim_{ x\rightarrow 1}\frac{16}{2}=\lim_{ x\rightarrow 1}8=8\)


If someone can point me in the right direction, I'd appreciate it. If I'm already on track, great!


~Eric

 

[Deleted member 4993]

I have a homework problem that I can't quite figure out:

Give an example of functions \(\displaystyle y=f(x)\) and \(\displaystyle y=g(x)\) such that \(\displaystyle \lim_{ x\rightarrow 1}f(x)=0\), but \(\displaystyle \lim_{ x\rightarrow 1} \frac{f(x)}{g(x)}=8\).

So far, the only functions that I have been able to conjure up are these:

\(\displaystyle f(x)=16x-16\)
\(\displaystyle g(x)=x^2-1\)

so

\(\displaystyle \lim_{x\rightarrow 1} \frac{f(x)}{g(x)}=\lim_{ x\rightarrow 1}\frac{16x-16}{x^2-1}=\lim_{ x\rightarrow 1}\frac{16(x-1)}{(x+1)(x-1)}=\lim_{ x\rightarrow 1}\frac{16}{x+1}=\)

\(\displaystyle \lim_{ x\rightarrow 1}\frac{16}{2}=\lim_{ x\rightarrow 1}8=8\)


If someone can point me in the right direction, I'd appreciate it. If I'm already on track, great!


~Eric

Looks good to me.......

 

[eric_f]

Looks good to me.......

Is there a reasonably simple mathematical way to determine these functions, or is it a game of educated guessing?

I know I can use multiples of eight in the numerator and raise the power of x in the denominator for every multiple of eight. Is this the way to do it?

Such as:

\(\displaystyle \lim_{x\rightarrow 1} \frac{8x-8}{x-1}=8\)

\(\displaystyle \lim_{x\rightarrow 1} \frac{16x-16}{x^2-1}=8\)

\(\displaystyle \lim_{x\rightarrow 1} \frac{24x-24}{x^3-1}=8\)

\(\displaystyle \lim_{x\rightarrow 1} \frac{32x-32}{x^4-1}=8\)

and so on...

 

[DrPhil]

Looks good to me.......

Yes........

Even simpler, f(x) = 8(x- 1)
...................g(x) = x - 1

[It don't get any no more simpler than that.]

g(x) can be any function which includes a single factor of (x - 1)
Then f(x) = 8 g(x)

 

[eric_f]

Thanks for the confirmations. I realized I was probably over-thinking it the more I started recognizing the pattern. :mrgreen:

 

[Bob Brown MSEE]

I have a homework problem that I can't quite figure out:
Give an example of functions \(\displaystyle y=f(x)\) and \(\displaystyle y=g(x)\) such that \(\displaystyle \lim_{ x\rightarrow 1}f(x)=0\), but \(\displaystyle \lim_{ x\rightarrow 1} \frac{f(x)}{g(x)}=8\).
~Eric

Many functions go through h(0)=0 smoothly and can be moved to x=1.

sinx / x -> 8sin(x-1)/(x-1)
tanx / x -> 8tanx(x-1)/(x-1)


ANY function that goes through h(0)=K smoothly and can be moved to h(0)-K.
Then treated as above...
i.e. f(x)=h(x-1+a)

Example:
e^x -> 8( e^(x-1) - 1 )/(x-1)


So now you can probably see how to take ANY non-horizontal point on a smooth funtion and...
1) move to (1,0)
2) divide by x-1
3) multiply by scale factor so limit is 8


coup de grâce

Scale and shift any 2 non-horizontal smooth functions
Examples:
8sin(x-1)/ln(x)
8(x^x - 1)/sin(x-1)