To find the limit [tex]\(\lim _{x \rightarrow 2} \frac{x^{-3}-2^{-3}}{x-2}\)[/tex], let's solve it step-by-step.
### Step 1: Rewrite the limit expression
First, rewrite the expression to make it easier to handle:
[tex]\[
\lim _{x \rightarrow 2} \frac{x^{-3} - 2^{-3}}{x - 2}
\][/tex]
### Step 2: Substitute known values
Recognize that [tex]\(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\)[/tex]. Therefore, the expression becomes:
[tex]\[
\lim _{x \rightarrow 2} \frac{x^{-3} - \frac{1}{8}}{x - 2}
\][/tex]
### Step 3: Recognize the indeterminate form
If we directly substitute [tex]\(x = 2\)[/tex] into the limit, both the numerator and the denominator approach 0, creating an indeterminate form [tex]\(\frac{0}{0}\)[/tex]. We can apply L'Hôpital's Rule in such cases, which involves differentiating the numerator and denominator separately.
### Step 4: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator with respect to [tex]\(x\)[/tex]:
1. The numerator: [tex]\( x^{-3} - \frac{1}{8} \)[/tex],
- Differentiate [tex]\( x^{-3} \)[/tex]:
[tex]\[
\frac{d}{dx}[x^{-3}] = -3x^{-4}
\][/tex]
- The term [tex]\(\frac{1}{8}\)[/tex] is a constant and its derivative is 0.
So, the derivative of the numerator is:
[tex]\[
-3x^{-4}
\][/tex]
2. Differentiate the denominator [tex]\( x - 2 \)[/tex]:
[tex]\[
\frac{d}{dx}[x - 2] = 1
\][/tex]
### Step 5: Substitute the derivatives back into the limit
Now, using L'Hôpital's Rule, the limit becomes:
[tex]\[
\lim _{x \rightarrow 2} \frac{-3x^{-4}}{1} = \lim _{x \rightarrow 2} -3x^{-4}
\][/tex]
### Step 6: Evaluate the limit by direct substitution
Substitute [tex]\(x = 2\)[/tex] into [tex]\(-3x^{-4}\)[/tex]:
[tex]\[
-3 \cdot (2^{-4})
\][/tex]
[tex]\[
2^{-4} = \frac{1}{2^4} = \frac{1}{16}
\][/tex]
Thus, the limit evaluates to:
[tex]\[
-3 \cdot \frac{1}{16} = -\frac{3}{16}
\][/tex]
Therefore,
[tex]\[
\lim _{x \rightarrow 2} \frac{x^{-3}-2^{-3}}{x-2} = -\frac{3}{16}
\][/tex]
