Question

find \\( \lim_{x \to c^+} f(x) \\) for the given function and value of \\( c \\).
\\( f(x) = \left( \frac{7x}{x + 1} \
ight) \left( \frac{7x + 7}{x^2 + x} \
ight), \\, c = 7 \\)
\\( \lim_{x \to 7^+} \left( \frac{7x}{x + 1} \
ight) \left( \frac{7x + 7}{x^2 + x} \
ight) = \square \\)
(type an integer or a simplified fraction.)

Explanation:

Step1: Factor and Simplify

Factor the second fraction: $\frac{7x + 7}{x^2 + x}=\frac{7(x + 1)}{x(x + 1)}$. Cancel $x + 1$ (for $x
eq - 1$).
Now the function becomes $\frac{7x}{x + 1}\cdot\frac{7(x + 1)}{x(x + 1)}=\frac{49}{x + 1}$ (after canceling $x$ and $x + 1$ where possible).

Step2: Substitute Limit

Now find $\lim_{x
ightarrow7^{+}}\frac{49}{x + 1}$. Substitute $x = 7$ into $\frac{49}{x + 1}$.
$\frac{49}{7+1}=\frac{49}{8}$

Answer:

$\frac{49}{8}$