Question

use lhôpitals rule to evaluate \\( \lim _{x \
ightarrow 10} \frac{x - 10}{x^{2} - 100} \\). then determine the limit using limit laws and commonly known limits.
use lhôpitals rule to rewrite the given limit so that it is not an indeterminate form.
\\( \lim _{x \
ightarrow 10} \frac{x - 10}{x^{2} - 100}=\lim _{x \
ightarrow 10}\left(\frac{1}{2 x}\
ight) \\)
choose the limit equivalent to the given limit that can be evaluated using limit laws and commonly known limits.
\\( \bigcirc \mathrm{a} \\) \\( \lim _{x \
ightarrow 10} \frac{1}{x - 10} \\)
\\( \bigcirc \mathrm{b} \\) \\( \lim _{x \
ightarrow 10} \frac{x}{x^{2}} \\)
\\( \bigcirc \mathrm{c} \\) \\( \lim _{x \
ightarrow 10} \frac{1}{x + 10} \\)
\\( \bigcirc \mathrm{d} \\) \\( \lim _{x \
ightarrow 10} \frac{\frac{1}{x}-\frac{10}{x^{2}}}{1 - \frac{100}{x^{2}}} \\)

Explanation:

Step1: Factor the denominator

We know that \(x^{2}-100=(x - 10)(x + 10)\). So, \(\lim_{x
ightarrow10}\frac{x - 10}{x^{2}-100}=\lim_{x
ightarrow10}\frac{x - 10}{(x - 10)(x + 10)}\)

Step2: Simplify the function

Cancel out the common factor \((x - 10)\) (for \(x
eq10\)). We get \(\lim_{x
ightarrow10}\frac{x - 10}{(x - 10)(x + 10)}=\lim_{x
ightarrow10}\frac{1}{x + 10}\)

Answer:

C. \(\lim_{x
ightarrow10}\frac{1}{x + 10}\)