Question

1. evaluate

(a) \\(\lim_{x \to 0} \frac{x + 7}{x^2 + 9x + 14}\\)
(b) \\(\lim_{x \to 3} \frac{x + 7}{x^2 + 9x + 14}\\)
(c) \\(\lim_{x \to -4} \frac{x + 7}{x^2 + 9x + 14}\\)
(d) \\(\lim_{x \to -7} \frac{x + 7}{x^2 + 9x + 14}\\)

Explanation:

Simplify the rational expression

$$ f(x) = \frac{x+7}{x^2+9x+14} = \frac{x+7}{(x+7)(x+2)} $$

For \(x
eq -7\):

$$ f(x) = \frac{1}{x+2} $$

Evaluate limits (a), (b), and (c) by direct substitution

For \(x \to 0\), \(x \to 3\), and \(x \to -4\), the function is continuous:

$$ \text{(a) } \lim_{x\to0} \frac{1}{x+2} = \frac{1}{0+2} = \frac{1}{2} $$

$$ \text{(b) } \lim_{x\to3} \frac{1}{x+2} = \frac{1}{3+2} = \frac{1}{5} $$

$$ \text{(c) } \lim_{x\to-4} \frac{1}{x+2} = \frac{1}{-4+2} = -\frac{1}{2} $$

Evaluate limit (d) using the simplified form

For \(x \to -7\), we resolve the \(\frac{0}{0}\) indeterminate form:

$$ \text{(d) } \lim_{x\to-7} \frac{x+7}{(x+7)(x+2)} = \lim_{x\to-7} \frac{1}{x+2} = \frac{1}{-7+2} = -\frac{1}{5} $$

Answer:

Question 6

(a) \(\frac{1}{2}\)

(b) \(\frac{1}{5}\)

(c) \(-\frac{1}{2}\)

(d) \(-\frac{1}{5}\)