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If k and b are constant such that lim x approach infinity (kx+b-(x^3+1)/(x^2+1)=0. Find the values of k and b

If k and b are constant such that lim x approach infinity (kx+b-(x^3+1)/(x^2+1)=0. Find-example-1
User Gwvatieri
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1 Answer

20 votes
20 votes

Combining the terms into one fraction, we have


kx + b - (x^3+1)/(x^2+1) = ((k-1)x^3 + bx^2 + kx + b - 1)/(x^2+1)

If this converges to 0 as
x\to\infty, then the degree of the numerator must be smaller than the degree of the denominator.

To ensure this, take
k=1 and
b=0. This eliminates the cubic and quadratic terms in the numerator, and we do have


\displaystyle \lim_(x\to\infty) (x - 1)/(x^2 + 1) = \lim_(x\to\infty) \frac{\frac1x - \frac1{x^2}}{1 + \frac1{x^2}} = 0

Alternatively, we can compute the quotient and remainder of the rational expression.


(x^3+1)/(x^2+1) = x - (x-1)/(x^2+1)

Then in the limit, we have


\displaystyle \lim_(x\to\infty) \left(kx + b - x + (x-1)/(x^2+1)\right) = (k-1) \lim_(x\to\infty) x + b = 0

Both terms on the left vanish if
k=1 and
b=0.

User Obo
by
2.8k points
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