Asked by John
How do I find
lim x-> infinity
√(x^2 + x) - √(x^2 - x)
lim x-> infinity
√(x^2 + x) - √(x^2 - x)
Answers
Answered by
MathMate
lim x->∞ (√(x^2 + x) - √(x^2 - x) )
We take advantage of the product:
(x+a)(x-a)=x²-a²
Multiply numerator and denominator by
(√(x^2 + x) + √(x^2 - x) )
to get
((√(x^2 + x) + √(x^2 - x) )(√(x^2 + x) - √(x^2 - x) ))/((√(x^2 + x) + √(x^2 - x) ))
=((x²+x)-(x²-x))/((√(x^2 + x) + √(x^2 - x) ))
=2x/((√(x^2 + x) + √(x^2 - x) ))
As x approaches infinity, the term x in the square-root radicals become insignificant and can be dropped out. So the denominator becomes
√(x²)+√(x²)
=2x
Therefore
lim x->∞ (√(x^2 + x) - √(x^2 - x) )
= 2x/2x
= 1
We take advantage of the product:
(x+a)(x-a)=x²-a²
Multiply numerator and denominator by
(√(x^2 + x) + √(x^2 - x) )
to get
((√(x^2 + x) + √(x^2 - x) )(√(x^2 + x) - √(x^2 - x) ))/((√(x^2 + x) + √(x^2 - x) ))
=((x²+x)-(x²-x))/((√(x^2 + x) + √(x^2 - x) ))
=2x/((√(x^2 + x) + √(x^2 - x) ))
As x approaches infinity, the term x in the square-root radicals become insignificant and can be dropped out. So the denominator becomes
√(x²)+√(x²)
=2x
Therefore
lim x->∞ (√(x^2 + x) - √(x^2 - x) )
= 2x/2x
= 1
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