Evaluate the following limits: 1. lim(x-3) (x3 - 27)/(x2 - 9) 2. lim(x-inf) (4x2 - 5x + 1)/sqrt(9x4 + 2x2…

Question

Evaluate the following limits: 1. lim(x->3) (x3 - 27)/(x2 - 9) 2. lim(x->inf) (4x2 - 5x + 1)/sqrt(9x4 + 2x2 - 6)

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the limit problems visible in your image.

$$
\frac{3^3 - 27}{3^2 - 9} = \frac{27 - 27}{9 - 9} = \frac{0}{0}
$$
This is an indeterminate form, so we need to simplify the expression.

Factor the numerator using the difference of cubes formula $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ and the denominator using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ .
$$ x^3 - 27 = (x-3)(x^2 + 3x + 9) $$ $$ x^2 - 9 = (x-3)(x+3) $$
Rewrite the limit with the factored expressions and cancel out the common term $(x-3)$ .
$$ \lim_{x \to 3} \frac{(x-3)(x^2 + 3x + 9)}{(x-3)(x+3)} = \lim_{x \to 3} \frac{x^2 + 3x + 9}{x+3} $$

$$
\frac{3^2 + 3(3) + 9}{3+3} = \frac{9 + 9 + 9}{6} = \frac{27}{6}
$$

Simplify the fraction.
$$ \frac{27}{6} = \frac{9}{2} $$ The limit is $\boxed{\frac{9}{2}}$.
2. We need to evaluate the limit: $\lim_{x \to \infty} \frac{4x^2 - 5x + 1}{\sqrt{9x^4 + 2x^2 - 6}}$
Identify the highest power of $x$ in the denominator.
Inside the square root, the highest power is $x^4$ , so $\sqrt{x^4} = x^2$ .
Divide

Evaluate the following limits: 1. lim(x->3) (x3 - 27)/(x2 - 9) 2. lim(x->inf) (4x2 - 5x + 1)/sqrt(9x4 + 2x2 - 6)