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compute the following two limits: \\(\\lim_{x \\to 5} \\frac{x^2 + 2x +…

Question

compute the following two limits:

\\(\lim_{x \to 5} \frac{x^2 + 2x + 1}{x^2 - 3x + 10}\\)

\\(\lim_{x \to 2} \frac{x^2 + 4x - 12}{\sqrt{x + 14} - 4}\\)

Explanation:

Evaluate the first limit by direct substitution

We check if direct substitution works.

$$ \lim_{x \to 5} \frac{x^2 + 2x + 1}{x^2 - 3x + 10} $$

Substitute \(x = 5\) into the numerator and denominator.

$$ \text{Numerator: } 5^2 + 2(5) + 1 = 25 + 10 + 1 = 36 $$
$$ \text{Denominator: } 5^2 - 3(5) + 10 = 25 - 15 + 10 = 20 $$

Since the denominator is non-zero, we evaluate directly.

$$ \frac{36}{20} = \frac{9}{5} $$

Analyze the second limit for indeterminate form

We check the second limit.

$$ \lim_{x \to 2} \frac{x^2 + 4x - 12}{\sqrt{x + 14} - 4} $$

Substitute \(x = 2\) to check the form.

$$ \text{Numerator: } 2^2 + 4(2) - 12 = 4 + 8 - 12 = 0 $$
$$ \text{Denominator: } \sqrt{2 + 14} - 4 = \sqrt{16} - 4 = 0 $$

This is an indeterminate form of type \(\frac{0}{0}\).

Simplify the second limit using rationalization

We multiply the numerator and denominator by the conjugate.

$$ \frac{x^2 + 4x - 12}{\sqrt{x + 14} - 4} \cdot \frac{\sqrt{x + 14} + 4}{\sqrt{x + 14} + 4} $$

Simplify the denominator.

$$ (\sqrt{x + 14} - 4)(\sqrt{x + 14} + 4) = (x + 14) - 16 = x - 2 $$

Factor the quadratic expression in the numerator.

$$ x^2 + 4x - 12 = (x - 2)(x + 6) $$

Combine these simplified parts.

$$ \frac{(x - 2)(x + 6)(\sqrt{x + 14} + 4)}{x - 2} $$

Cancel common factors and compute the limit

Cancel the common factor \((x - 2)\) for \(x
eq 2\).

$$ (x + 6)(\sqrt{x + 14} + 4) $$

Now, evaluate the limit by direct substitution.

$$ \lim_{x \to 2} (x + 6)(\sqrt{x + 14} + 4) = (2 + 6)(\sqrt{2 + 14} + 4) $$
$$ = 8 \cdot (4 + 4) = 8 \cdot 8 = 64 $$

Answer:

Question 1

$$ \lim_{x \to 5} \frac{x^2 + 2x + 1}{x^2 - 3x + 10} = \frac{9}{5} $$

Question 2

$$ \lim_{x \to 2} \frac{x^2 + 4x - 12}{\sqrt{x + 14} - 4} = 64 $$