## Need help proving limit does not exist!

 Quote by SammyS I have fixed your Latex above, and repeat it below, so you can see the code. $(|x|^{a+(bc/d)}) /(2(|x|^c))$ Now as a more readable fraction: $\displaystyle \frac{|x|^{a+(bc/d)}}{2(|x|^c)}$ Write that expression using |x| only once. Remember, $\displaystyle \frac{x^R}{x^T}=x^{R-T}\ .$
I got this after: $(|x|^{a+(bc/d)-c}) /{2}$
expanding the exponents I have $\displaystyle \frac{acd+bc^2-c^2d}{cd}$
and given $\displaystyle \frac{ad+bc-cd}{cd}<=0$ which is 1/c of the exponent..
So then |x| has an exponent negative so limit d.n.e right? cause limit of 1/|x| as x->0 d.n.e

Mentor
 Quote by mathie_geek I got this after: $(|x|^{a+(bc/d)-c}) /{2}$ expanding the exponents I have $\displaystyle \frac{acd+bc^2-c^2d}{cd}$ and given $\displaystyle \frac{ad+bc-cd}{cd}<=0$ which is 1/c of the exponent.. So then |x| has an exponent negative so limit d.n.e right? cause limit of 1/|x| as x->0 d.n.e
Actually, that's $\displaystyle \frac{ad+bc-cd}{d}\le 0$

What if $\displaystyle \frac{ad+bc-cd}{d}=0\ ?$

Now, maybe you can adjust the path so that the exponent is always zero.

 Quote by SammyS Actually, that's $\displaystyle \frac{ad+bc-cd}{d}\le 0$ What if $\displaystyle \frac{ad+bc-cd}{d}=0\ ?$ Now, maybe you can adjust the path so that the exponent is always zero.
Oh, I see. Since $\displaystyle \frac{ad+bc-cd}{d}> \displaystyle \frac{ad+bc-cd}{cd}$ then does that mean it HAS to = 0? If it's 0 then lim is 1/2

Mentor
 Quote by mathie_geek ... expanding the exponents I have $\displaystyle \frac{acd+bc^2-c^2d}{cd}$ and given $\displaystyle \frac{ad+bc-cd}{cd}<=0$ which is 1/c of the exponent.. ...
Let's go back to the exponent:
$\displaystyle \frac{acd+bc^2-c^2d}{cd}=\frac{c(ad+bc-cd)}{cd}=\frac{ad+bc-cd}{d}\ .$
I have no idea as to why you multiplied that by 1/c .

 Quote by mathie_geek Oh, I see. Since $\displaystyle \frac{ad+bc-cd}{d}> \displaystyle \frac{ad+bc-cd}{cd}$ then does that mean it HAS to = 0? If it's 0 then lim is 1/2
Unless you meant to say that "a,b,c,d are positive integers numbers" (emphasizing integers, rather than numbers), the following inequality
$\displaystyle \frac{ad+bc-cd}{d}>\frac{ad+bc-cd}{cd} \ \ \text{ is not true.}$
The number, c, might very well be less than 1 if a, b, c, and d are not required to be integers.

All that I said was, if a/c + b/d ≤ 1, then $\displaystyle \frac{ad+bc-cd}{d}\le 0$.

If you choose the case of it being an equality, (choose the = sign), then the limit is defined, with said limit being 1/2.

However, that's far from being the general case for this expression.

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