Is f(x) = (x-iy)/(x-1) a Continuous Function?

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The function f(x) = (x-iy)/(x-1) is analyzed for continuity, focusing on the behavior at x = 1 where the denominator becomes zero. Both u(x) = x-iy and v(x) = x-1 are continuous functions, but the overall function f(x) is not continuous at x = 1 due to the division by zero. The discussion highlights that while the limit may exist, the function itself is undefined at that point, indicating a discontinuity. The concept of removable discontinuities is mentioned, emphasizing that defining the function at the point of discontinuity could resolve the issue. Thus, f(x) is continuous everywhere except at x = 1.
Krayfish
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Homework Statement


Determine if the following function is continuous: f(x) = (x-iy)/(x-1)

Homework Equations


How do find out if a function is continuous without graphing it and without a point to examine? I know I've learned this, probably in pre-calculus too, but I'm blanking

The Attempt at a Solution


u(x) = x-iy as a function is continuous because, due to the i term, x-iy will never equal 0 and it is a linear function
v(x) = x-1 is also continuous,as it is a linear function that exists under all conditions

if u and v are both continuous under all conditions, than u/v must also be continuous?
 
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From the point of view of precalculus, you have learned about limits, in particular, one-sided limits. What happens at x = 1?
 
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Krayfish said:
if u and v are both continuous under all conditions, than u/v must also be continuous?
u/v will be continuous at points for which v ≠ 0...
 
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Informally, you can't divide by 0. The function may blow up to +-infinity there. At any rate, it's not defined when the denominator is 0.

Formally, you should see a theorem in your book that for f(x) and g(x) continuous, f(x)/g(x) is continuous except where g(x) = 0. The limit of f(x)/g(x) may exist as g(x)->0, for instance in the function sin(x) / x, but the function itself is not continuous there. (This is a fine point called a "removable discontinuity", which is also in your textbook. Basically, if the limit exists, then you define the function to say "and when the denominator = 0, it's defined to be the limit")
 
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Krayfish said:

Homework Statement


Determine if the following function is continuous: f(x) = (x-iy)/(x-1)
Did you mean ##f(z)##? If not, then how is ##y## defined?

u(x) = x-iy as a function is continuous because, due to the i term, x-iy will never equal 0 and it is a linear function
Does it matter if ##u## is ever equal to 0?
 
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verty said:
From the point of view of precalculus, you have learned about limits, in particular, one-sided limits. What happens at x = 1?
OH yeah that one was pretty obvious, that's what you get when you're doing math with no sleep. Undefined, thank you
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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