Compute the limit of (3x^3 - 2x - 1) / (5x^2 + 3x - 8) as x approaches 1.

Step-by-step solution

1. 1

   Substitute $x=1$ first. Numerator: $3-2-1=0$. Denominator: $5+3-8=0$. The form is $\tfrac00$, so the limit is indeterminate — but the shared zero at $x=1$ tells us $(x-1)$ divides both polynomials, which is the key to resolving it.

2. 2

   Factor $(x-1)$ out of the numerator. Synthetic division of $3x^3+0x^2-2x-1$ by $x-1$ gives coefficients $3,\ 3,\ 1$ with remainder $0$:

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

3. 3

   Factor $(x-1)$ out of the denominator.

   $5x^2+3x-8=(x-1)(5x+8)$

   Check by expanding: $5x^2+8x-5x-8=5x^2+3x-8\ \checkmark$.

4. 4

   Cancel and substitute. For $x\ne 1$ the common factor cancels, and the remaining expression is continuous at $x=1$:

   $\lim_{x\to 1}\frac{3x^2+3x+1}{5x+8}=\frac{3+3+1}{5+8}=\frac{7}{13}$

5. 5

   Cross-check with l'Hopital's rule and a numerical probe. Differentiating top and bottom gives $\dfrac{9x^2-2}{10x+3}$, which at $x=1$ is $\dfrac{7}{13}$ — the same value. Numerically, the original quotient at $x=1.0001$ is $0.5385101$, against $7/13=0.5384615$, consistent to four decimals.