Hi,
Can someone explain how to evaluate:
$\displaystyle \lim_{\Re (z)\to-\infty} \left|\frac{1+\frac{1}{2}z}{1-\frac{1}{2}z}\right|$
Would I need L'Hopital's rule?
I assume the imaginary part of z is remaining finite .....? $\displaystyle \lim_{x \rightarrow - \infty} \left| \frac{1 + \frac{1}{2} x + i \frac{1}{2} y}{1 - \frac{1}{2} x - i \frac{1}{2} y} \right| = \lim_{x \rightarrow - \infty} \left| \frac{\frac{1}{x} + \frac{1}{2} + i \frac{y}{2x}}{\frac{1}{x} - \frac{1}{2} - i \frac{y}{2x}}\right|$
Yes, I believe the Im(z) remains finite. I'm basically trying to understand how Wikipedia found the limit on this page: Stiff equation - Wikipedia, the free encyclopedia , section "Example: The Euler and trapezoidal methods".
So, about what you did: I assume $\displaystyle y$ is the imaginary part of z? Where did the $\displaystyle i$ vanish to?
But from what you did, it looks like the limit becomes: (since the other terms vanish).
$\displaystyle \lim_{x\to -\infty} \left|\frac{ +\frac{1}{2} }{- \frac{1}{2} }\right| = |-1| = 1$.
Which is what wikipedia got, but is my working right?
it's a basic algebra thing: if we have a quotient Q/P, where Q, P are polynomials on some unknow t of the same degree (on t), then we can always write Q/P = 1 - Q'/P, for some pol. in t of degree at most deg(Q)
After that all is simple limits: as it is x --> -oo all the rest is taken as constants, and thus (x + constant)/(-x + constant) --> -1
Tonio