Question

iii) $lim_{x
ightarrow - 5}\frac{x^{2}+6x + 5}{\frac{1}{5}+\frac{1}{x}}$

Explanation:

Step1: Simplify the denominator

First, find a common - denominator for the denominator $\frac{1}{5}+\frac{1}{x}$. The common denominator is $5x$, so $\frac{1}{5}+\frac{1}{x}=\frac{x + 5}{5x}$.

Step2: Rewrite the original limit

The original limit $\lim_{x
ightarrow - 5}\frac{x^{2}+6x + 5}{\frac{1}{5}+\frac{1}{x}}$ can be rewritten as $\lim_{x
ightarrow - 5}\frac{x^{2}+6x + 5}{\frac{x + 5}{5x}}=\lim_{x
ightarrow - 5}\frac{5x(x^{2}+6x + 5)}{x + 5}$.

Step3: Factor the numerator

Factor the quadratic expression $x^{2}+6x + 5=(x + 1)(x + 5)$. Then the limit becomes $\lim_{x
ightarrow - 5}\frac{5x(x + 1)(x + 5)}{x + 5}$.

Step4: Cancel out the common factor

Since $x
eq - 5$ (when taking the limit), we can cancel out the common factor $(x + 5)$ in the numerator and the denominator. So we get $\lim_{x
ightarrow - 5}5x(x + 1)$.

Step5: Substitute the value of $x$

Substitute $x=-5$ into $5x(x + 1)$. We have $5\times(-5)\times(-5 + 1)=5\times(-5)\times(-4)=100$.

Answer:

$100$