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| Math Central | Quandaries & Queries |
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Question from David, a student: Find the limit. (Hint: treat the expression as a fraction whose denominator is 1, and rationalize the numerator.) lim [x + squareroot(x^2 + 3)] as x->-inf i got to lim -3/(x - squareroot(x^2 + 3)) as x->-inf but i'm having trouble understanding why the answer is 0 plz explain thx |
David,
I think the best action at this point is to make a change of variables z = -x then the expression becomes
-3/(-z - sqrt(z2 + 3)) = 3/(z + sqrt(z2 + 3))
and the limit changes to z -> ∞. Now since z is approaching plus infinity I can assume z is positive so z = √(z2). Divide the numerator and denominator by z
3/(z + sqrt(z2 + 3)) = (3/z)/(1 + sqrt(1 + 3/z2))
Now as z -> ∞ the numerator approaches 0 and the denominator approaches 2. Thus the limit is 0.
Harley
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