Sin is the root of all evil
x → 0 + lim x sin ( x ) + sin ( x ) + sin ( x ) + … = ?
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3 solutions
But I think it should be f(x)=[1 - sqrt(1+4sinx)]/2 because putting x=0 in sqrt(sinx+sqrt(sinx+sqrt(......... gives you 0 whereas putting x=0 in your f(x) gives 1
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It's true that f ( 0 ) = 0 , but lim x → 0 + f ( x ) = 1 ; nobody claimed that f ( x ) is continuous at x = 0 . Your formula for f ( x ) gives negative values for small positive x , which is impossible for a (nested) square root.
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I don't understand, what do you mean about negative here? Please explain more. I don't see any negative value from his formula for f(x).
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@Hafizh Ahsan Permana – 2 1 − 1 + 4 sin x is negative for 0 < x < π , but f ( x ) cannot be negative, being a square root. Thats why we have f ( x ) = 2 1 + 1 + 4 sin x for 0 < x < π .
A bit agree to Hamza. Why do we use
f ( x ) = [ 1 + s q r t ( 1 + 4 s i n x ) ] / 2
Not f ( x ) = [ 1 − s q r t ( 1 + 4 s i n x ) ] / 2
Well the result is the same at the end. The limit is does'nt exist. Correct me if i wrong.
First step,may be, is just to verify this expression s i n x + s i n x + s i n x . . . let y = s i n x + s i n x + s i n x . . . = s i n x + y then y = 2 1 + 1 + 4 s i n x ≈ 1 + s i n x So, the value is l i m x − > 0 x s i n x + s i n x + s i n x . . . = x 1 + s i n x since x approach zero from both side so, the limit does not exist.
Moderator note:
Nice! Although it's better to explain why 2 1 + 1 + 4 sin x ≈ 1 + sin x for x ≈ 0 .
Ok i'll explain that part
2 1 + ( 1 + 4 s i n x ) 2 1
Since x → 0 ⟹ s i n x → 0
We can apply binomial theorem here
( 1 + x ) n = 1 + n x , x ≪ 1
Our equation reduces to
2 1 + ( 1 + 2 4 s i n x ) = 1 + s i n x
1/0 isn't it undefined?
lim x->0+ (1/x)=+infinity; lim x->0- (1/x)= -infinity; therefore limit at x=0 is does not exist for the given function
I think there's an error in your reasoning, but I can't see your sheet clearly.
I agree with you up to the line y^2 - y - sin(x) = 0.
From here, the quadratic formula should yield y = (1/2)(-1 +/- sqrt(1 + 4*sin(x))), where we must take the plus sign since y must be positive.
So y = -1/2 + (1/2)(1 + 4*sin(x)) = sin(x) + o(x) using the series expansion for the square root.
The desired limit is 1 by my reckoning.
Curious to know if I missed anything.
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You are making a trivial sign error: It's y = 2 1 + . . . by the quadratic formula.
How can you claim that lim x → 0 − . . . . = − ∞ ? The numerator is undefined in that case!
The limit fails to exist because the numerator is undefined for negative x with − π < x < 0 .
Out of curiosity, let's find the single-sided limit, for positive x. Define f ( x ) = sin x + ( . . . ) for 0 < x < π . Letting f ( x ) = sin x + f ( x ) we find that f ( x ) = 2 1 + 1 + 4 sin x . Now lim x → 0 + x f ( x ) = ∞ since the numerator goes to 1 and the denominator goes to 0.