Lim as x approaches infinity square root of 4x^2 + 1/ (x-1)
The sq root is only over the 4x^2 + 1
The sq root is only over the 4x^2 + 1
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√(4x^2 + 1) / (x-1)
divide everything by x.
√(4x^2 + 1)/x / (x-1)/x
√(4x^2/x^2 + 1/x^2) / (x/x - 1/x)
√(4 + 1/x^2) / (1 - 1/x)
as x--> ∞
√(4 + 0) / (1 - 0)
√4/1
2/1 = 2
divide everything by x.
√(4x^2 + 1)/x / (x-1)/x
√(4x^2/x^2 + 1/x^2) / (x/x - 1/x)
√(4 + 1/x^2) / (1 - 1/x)
as x--> ∞
√(4 + 0) / (1 - 0)
√4/1
2/1 = 2
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Use L'Hopital's rule
Derivative of numerator = 4x/(4x^2+1)^(1/2)
Derivative of denominator = 1
lim x-> inf 4x/(4x^2+1)^(1/2) = 4x/2x = 2
Derivative of numerator = 4x/(4x^2+1)^(1/2)
Derivative of denominator = 1
lim x-> inf 4x/(4x^2+1)^(1/2) = 4x/2x = 2
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At the limit, we can forget about the +1, so we get lim (x -> inf) sqrt 4x^2 / (x - 1) = lim (x -> inf) 2x / (x - 1), which is 2.
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The answer is 2.