Question

problem. 5 : determine if the limit approaches a finite number, $\infty$, $-\infty$, or does not exist. (if the limit does not exist, write dne)
$\lim\limits_{x \to 5} \frac{x^2 - 7x + 10}{x^2 - 25} = $
problem. 6 : determine if the limit approaches a finite number, $\infty$, $-\infty$, or does not exist. (if the limit does not exist, write dne)
$\lim\limits_{x \to -2} \frac{-3x^3 - 24}{x^2 + x - 2} = $
problem. 7 : calculate the following limit:
$\lim\limits_{x \to -\sqrt{3}} \frac{8(x^4 - 9)}{x^2 - 3} = $
problem. 8 : compute the following limit:
$\lim\limits_{x \to -8} -\frac{4(x + 5)^2 + 1}{\frac{1}{3x + 2} + 2} = $

Explanation:

Problem 5

Step1: Factor numerator and denominator

Numerator: \(x^2 - 7x + 10=(x - 2)(x - 5)\)
Denominator: \(x^2 - 25=(x - 5)(x + 5)\)
So the limit becomes \(\lim_{x\to 5}\frac{(x - 2)(x - 5)}{(x - 5)(x + 5)}\)

Step2: Cancel common factor \((x - 5)\) (for \(x

eq5\))
We get \(\lim_{x\to 5}\frac{x - 2}{x + 5}\)

Step3: Substitute \(x = 5\)

\(\frac{5 - 2}{5 + 5}=\frac{3}{10}\)

Step1: Factor denominator

Denominator: \(x^2 + x - 2=(x + 2)(x - 1)\)
So the limit is \(\lim_{x\to -2}\frac{-3x^3 - 24}{(x + 2)(x - 1)}\)

Step2: Factor numerator (using \(a^3 + b^3=(a + b)(a^2 - ab + b^2)\))

Numerator: \(-3x^3 - 24=-3(x^3 + 8)=-3(x + 2)(x^2 - 2x + 4)\)
Now the limit becomes \(\lim_{x\to -2}\frac{-3(x + 2)(x^2 - 2x + 4)}{(x + 2)(x - 1)}\)

Step3: Cancel common factor \((x + 2)\) (for \(x

eq -2\))
We get \(\lim_{x\to -2}\frac{-3(x^2 - 2x + 4)}{x - 1}\)

Step4: Substitute \(x = -2\)

\(\frac{-3((-2)^2 - 2(-2) + 4)}{-2 - 1}=\frac{-3(4 + 4 + 4)}{-3}=\frac{-3\times12}{-3}=12\)

Step1: Factor numerator (using \(a^2 - b^2=(a - b)(a + b)\))

Numerator: \(x^4 - 9=(x^2)^2 - 3^2=(x^2 - 3)(x^2 + 3)\)
So the limit is \(\lim_{x\to -\sqrt{3}}\frac{8(x^2 - 3)(x^2 + 3)}{x^2 - 3}\)

Step2: Cancel common factor \((x^2 - 3)\) (for \(x

eq\pm\sqrt{3}\))
We get \(\lim_{x\to -\sqrt{3}}8(x^2 + 3)\)

Step3: Substitute \(x = -\sqrt{3}\)

\(8((-\sqrt{3})^2 + 3)=8(3 + 3)=8\times6 = 48\)

Answer:

\(\frac{3}{10}\)

Problem 6