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Complex Limits (1 Viewer)
- Thread starter mojo1
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Obvious
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- Mar 22, 2010
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- HSC
- 2013
- Uni Grad
- 2016
Considering that it's a continuous function the limit would be f(2) = 6
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To prove this using the epsilon-delta definition we have to show that for all M > 0 there exists some N > 0 such that |f(z) - f(2)| < N for all z in the complex plane with |z - 2| < M.
|f(z) - f(2)| = |z^2 + z - 6| = |z^2 - 4z + 4 + 5z - 10| <= |(z - 2)^2| + |5(z - 2)|* < M^2 + 5M
*Triangle inequality
So N = M^2 + 5M and N approaches zero as M does so that's fine.
_________________
To find the derivative we take the limit of the difference quotient d(h) as h approaches zero (d(h) = f(z + h) - f(z)/h) . So just substitute the relevant values, take the limit and you should get that it (which is the derivative) exists everywhere in C and equals 2z + 1.
_________________
To prove this using the epsilon-delta definition we have to show that for all M > 0 there exists some N > 0 such that |f(z) - f(2)| < N for all z in the complex plane with |z - 2| < M.
|f(z) - f(2)| = |z^2 + z - 6| = |z^2 - 4z + 4 + 5z - 10| <= |(z - 2)^2| + |5(z - 2)|* < M^2 + 5M
*Triangle inequality
So N = M^2 + 5M and N approaches zero as M does so that's fine.
_________________
To find the derivative we take the limit of the difference quotient d(h) as h approaches zero (d(h) = f(z + h) - f(z)/h) . So just substitute the relevant values, take the limit and you should get that it (which is the derivative) exists everywhere in C and equals 2z + 1.
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