Question

If $\lim\limits_{x\to\infty} (\frac{x^3+1}{x+1}-ax^2-bx-c)=4$ where $a,b,c \in R$, then, the value of $|a+b+c|$ is:

Answer 1

3

Answer 2

To evaluate the given limit and find the values of $a, b, c$, we first simplify the expression inside the limit.

The given limit is: $\lim\limits_{x\to\infty} \left(\frac{x^3+1}{x+1}-ax^2-bx-c\right)=4$

Step 1: Simplify the rational expression $\frac{x^3+1}{x+1}$. We know the sum of cubes factorization: $x^3+1 = (x+1)(x^2-x+1)$. So, for $x \neq -1$: $\frac{x^3+1}{x+1} = \frac{(x+1)(x^2-x+1)}{x+1} = x^2-x+1$

Step 2: Substitute the simplified expression back into the limit. $\lim\limits_{x\to\infty} (x^2-x+1 - ax^2 - bx - c) = 4$

Step 3: Group the terms by powers of $x$. $\lim\limits_{x\to\infty} ((1-a)x^2 + (-1-b)x + (1-c)) = 4$

Step 4: Determine the values of $a, b, c$ for the limit to be finite. For the limit of a polynomial as $x \to \infty$ to be a finite value, all terms with powers of $x$ greater than 0 must vanish (i.e., their coefficients must be zero). Otherwise, the limit would be $\pm\infty$.

Therefore, the coefficient of $x^2$ must be zero: $1-a = 0 \implies a=1$

Now, substitute $a=1$ back into the expression: $\lim\limits_{x\to\infty} (0 \cdot x^2 + (-1-b)x + (1-c)) = 4$ $\lim\limits_{x\to\infty} ((-1-b)x + (1-c)) = 4$

Next, the coefficient of $x$ must be zero: $-1-b = 0 \implies b=-1$

Substitute $b=-1$ back into the expression: $\lim\limits_{x\to\infty} (0 \cdot x + (1-c)) = 4$ $\lim\limits_{x\to\infty} (1-c) = 4$

Finally, the constant term must be equal to the value of the limit: $1-c = 4 \implies c = 1-4 \implies c=-3$

Step 5: Calculate the value of $|a+b+c|$. We have $a=1$, $b=-1$, and $c=-3$. $a+b+c = 1 + (-1) + (-3) = 1 - 1 - 3 = -3$ $|a+b+c| = |-3| = 3$