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

Step-by-step solution

1. 1

   Substitute first to see what kind of limit this is. Direct substitution is always the first move:

   $3(1)^3-2(1)-1=0,\qquad 5(1)^2+3(1)-8=0$

   The result is $\tfrac{0}{0}$, which is not an answer — it is a signal that the two polynomials share a factor.

2. 2

   Use the factor theorem. A polynomial $p$ with $p(1)=0$ is divisible by $x-1$. Both of ours vanish at $x=1$, so both carry that factor. Dividing the numerator by $x-1$:

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

   Expanding back confirms it: $3x^3+3x^2+x-3x^2-3x-1=3x^3-2x-1$.

3. 3

   Factor the denominator the same way.

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

   Check: $5x^2+8x-5x-8=5x^2+3x-8$. Both factorizations are exact, so nothing has been approximated.

4. 4

   Cancel — and note why that is legal. A limit as $x\to 1$ only looks at $x\neq 1$, and for $x\neq 1$ the factor $x-1$ is a nonzero number you may divide out:

   $\frac{(x-1)(3x^2+3x+1)}{(x-1)(5x+8)}=\frac{3x^2+3x+1}{5x+8}$

   The new function differs from the old one only at the single point $x=1$, which the limit ignores.

5. 5

   Substitute again into the reduced fraction. The denominator no longer vanishes ($5+8=13\neq 0$), so substitution now works:

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

6. 6

   Sanity check numerically. At $x=1.001$ the original quotient is $0.53895$ and at $x=0.999$ it is $0.53798$; both bracket $\tfrac{7}{13}=0.53846$, so the value is confirmed from both sides.