Question

in the following exercises, use direct substitution to show that each limit leads to the indeterminate form 0/0. then, evaluate the limit.

1. \\( \lim _ { x \

ightarrow 4 } \frac { x ^ { 2 } - 16 } { x - 4 } \\)

1. \\( \lim _ { x \

ightarrow 2 } \frac { x - 2 } { x ^ { 2 } - 2 x } \\)

Explanation:

Problem 93

Step1: Direct substitution

Substitute \(x = 4\) into \(\frac{x^{2}-16}{x - 4}\).
For the numerator: \(x^{2}-16=(4)^{2}-16=16 - 16=0\).
For the denominator: \(x - 4=4 - 4=0\). So, we get the indeterminate form \(\frac{0}{0}\).

Step2: Simplify the function

Factor the numerator \(x^{2}-16=(x + 4)(x - 4)\). Then \(\frac{x^{2}-16}{x - 4}=\frac{(x + 4)(x - 4)}{x - 4}=x + 4\) (for \(x
eq4\)).

Step3: Evaluate the limit

Now, find \(\lim_{x
ightarrow4}(x + 4)\). Substitute \(x = 4\) into \(x+4\), we get \(4 + 4=8\).

Step1: Direct substitution

Substitute \(x = 2\) into \(\frac{x - 2}{x^{2}-2x}\).
For the numerator: \(x - 2=2 - 2=0\).
For the denominator: \(x^{2}-2x=(2)^{2}-2\times2=4 - 4=0\). So, we get the indeterminate form \(\frac{0}{0}\).

Step2: Simplify the function

Factor the denominator \(x^{2}-2x=x(x - 2)\). Then \(\frac{x - 2}{x^{2}-2x}=\frac{x - 2}{x(x - 2)}=\frac{1}{x}\) (for \(x
eq2\)).

Step3: Evaluate the limit

Now, find \(\lim_{x
ightarrow2}\frac{1}{x}\). Substitute \(x = 2\) into \(\frac{1}{x}\), we get \(\frac{1}{2}\).

Answer:

\(\lim_{x
ightarrow4}\frac{x^{2}-16}{x - 4}=8\)

Problem 94