Question

a. $\lim\limits_{x \to 2} x^3 - 6x + 3 =$
b. $\lim\limits_{x \to 1} \frac{x^2 + 1}{3x^2} =$

Explanation:

Part a

Step1: Substitute \( x = 2 \) into the function

For the limit \( \lim_{x \to 2} (x^3 - 6x + 3) \), since the function \( f(x)=x^3 - 6x + 3 \) is a polynomial, we can use the direct substitution property of limits for polynomials. Substitute \( x = 2 \) into the function:
\( f(2)=2^3-6\times2 + 3 \)

Step2: Calculate the value

First, calculate \( 2^3=8 \), \( 6\times2 = 12 \). Then:
\( 8-12 + 3=-1 \)

Step1: Substitute \( x = 1 \) into the function

For the limit \( \lim_{x \to 1}\frac{x^2 + 1}{3x^2} \), the function \( g(x)=\frac{x^2 + 1}{3x^2} \) is a rational function, and at \( x = 1 \), the denominator \( 3x^2=3\times1^2 = 3
eq0 \), so we can use direct substitution. Substitute \( x = 1 \) into the function:
\( g(1)=\frac{1^2 + 1}{3\times1^2} \)

Step2: Calculate the value

Calculate the numerator \( 1^2+1 = 2 \), the denominator \( 3\times1^2=3 \). Then:
\( \frac{2}{3} \)

Answer:

\( -1 \)

Part b