Are you in your Limits?

Calculus Level 3

$\large \lim_{x\to\infty} \left( \frac{x^3+1}{x^2 + 1} - (ax+b) \right) = 2$

If the limit above is true for constants $a$ and $b$ , find the value of $b$ .

2 solutions

Raj Rajput
Sep 16, 2015

nice handwriting

Ponhvoan Srey - 5 years, 8 months ago

thank you :)

RAJ RAJPUT - 5 years, 8 months ago

Gian Sanjaya
Sep 16, 2015

Notice that:

$\frac{x^3+1}{x^2+1}-(ax+b)=x-\frac{x-1}{x^2+1}-ax-b=(1-a)x-(b+\frac{x-1}{x^2+1})$

Also, since the degree of x is less than the degree of x^2:

$\lim_{x->\infty} \frac{x-1}{x^2+1} = 0$

while $\lim_{x->\infty} x = \infty$ . As a result, we have $1-a=0$ and we have $-b=2$ ; $b = \boxed{-2}$ .