solve it quickly it's urgent no spam answer plz
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Answer:
k = 0
Step-by-step explanation:
You need
lim (x->0) f(x) = f(0) = k
Whatever k is, the numerator in f(x), for x≠0, is a constant. So if the numerator is not zero, then f(x) -> ∞ when x->0. This is no good. So the numerator must be 0.
That means we must have 1-cos(4k) = 0, and this makes f(x) = 0 for all x≠0. Therefore
lim (x->0) f(x) = lim (x->0) 0 = 0.
So k = 0.
Checking that this does indeed make the numerator 0:
1 - cos(4k) = 1 - cos 0 = 1 - 1 = 0.... Good!
Anonymous:
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but ans is 1/2
The answer is deinitely not 1/2! That would result in f(x) -> infinity as x->0, and that means that f(x) would not be continuous at x=0 in that case. The solution above already explains the details so there can be no doubt.
Perhaps there is a typo in the question. If instead of "4k" you put "4x" and instead of "8x^2" you put "16x^2", then the answer would be 1/2. But that takes a bit more work to show it!
Perhaps there is a typo in the question. If instead of "4k" you put "4x" and instead of "8x^2" you put "16x^2", then the answer would be 1/2. But that takes a bit more work to show it!
this is a wrong question and answer.
The answer is correct for the question posed. If there is an error in the question, that's too bad, although the comment above does offer suggestions for what the intended question might have been... but we can't be sure!
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