Given that f(x) is a function continuous at x=1 and satisfy the functional equation f(xy)=f(x)f(y) for all x and y. Prove that f(x) is continuous at all non-zero x.
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f(1)2=f(1)⇒f(1)=1,or f(1)=0.f(1)=0⇒f(x)=0∀x∈R⇒f is continuous over R. Otherwise, f(1)=1.Suppose we had a sequence of real numbers, (an), such that, limn→∞(an)=1.The existence of such a sequence is guaranteed by the density of real numbers.Now f is continuous at 1, so, by the definition of continuity, limn→∞f((an))=f(limn→∞(an))=f(1)=1.Take an arbitrary real number, c. Now, limn→∞(c⋅an)=c⋅limn→∞(an)=c.So, f(c)=f(c)⋅1=f(c)⋅(limn→∞f(an))=limn→∞f(c)⋅f(an)=limn→∞f(c⋅(an)).But, limn→∞c⋅(an)=c.So, as n→∞,c⋅an→c. Set c⋅an=x. Then, limx→cf(x)=f(c), which is what we needed to show. This finishes the proof! I like your questions, Akhilesh.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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f(1)2=f(1)⇒f(1)=1,or f(1)=0.f(1)=0⇒f(x)=0 ∀ x∈R⇒f is continuous over R. Otherwise, f(1)=1. Suppose we had a sequence of real numbers, (an), such that, limn→∞(an)=1.The existence of such a sequence is guaranteed by the density of real numbers.Now f is continuous at 1, so, by the definition of continuity, limn→∞f((an))=f(limn→∞(an))=f(1)=1. Take an arbitrary real number, c. Now, limn→∞(c⋅an)=c⋅limn→∞(an)=c.So, f(c)=f(c)⋅1=f(c)⋅(limn→∞f(an))=limn→∞f(c)⋅f(an)=limn→∞f(c⋅(an)).But, limn→∞c⋅(an)=c.So, as n→∞,c⋅an→c. Set c⋅an=x. Then, limx→cf(x)=f(c), which is what we needed to show. This finishes the proof! I like your questions, Akhilesh.
Can you please provide a solution for this one too.@Rishabh Cool
e raised to x
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I dont think ex can be the answer as it is continuous at x=0 too. Please a provide a solution if you have arrived at this answer.
@parv morCan you please post a solution to this one too