Evaluate the limit using L'Hospital's rule $$ \lim _{x \rightarrow 0} \frac{8^{x}-9^{x}}{x} $$

Question

UpStudy AI · Accepted Answer

The given limit is

$$
\lim _{x \rightarrow 0} \frac{8^{x}-9^{x}}{x}.
$$

Step 1: Notice that when $$ x = 0 $$, both the numerator and the denominator become zero:

$$
8^0 - 9^0 = 1 - 1 = 0 \quad \text{and} \quad x=0.
$$

Thus, the limit is in the indeterminate form $$ \frac{0}{0} $$.

Step 2: Apply L'Hospital's Rule, which states that for functions $$ f(x) $$ and $$ g(x) $$ with

$$
\lim_{x \to c} f(x) = 0 \quad \text{and} \quad \lim_{x \to c} g(x)=0,
$$

we have

$$
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)},
$$

provided the limit on the right-hand side exists.

Step 3: Differentiate the numerator and denominator with respect to $$ x $$.

The derivative of the numerator is determined using the fact that

$$
\frac{d}{dx}a^x = a^x \ln a.
$$

Thus,

$$
\frac{d}{dx} \left( 8^x - 9^x \right) = 8^x \ln 8 - 9^x \ln 9.
$$

The derivative of the denominator is

$$
\frac{d}{dx} (x) = 1.
$$

Step 4: Substitute $$ x = 0 $$ into the differentiated expression:

$$
\lim _{x \rightarrow 0} \frac{8^{x} \ln 8 - 9^{x} \ln 9}{1} = 8^0 \ln 8 - 9^0 \ln 9.
$$

Since

$$
8^0 = 1 \quad \text{and} \quad 9^0 = 1,
$$

the expression simplifies to

$$
\ln 8 - \ln 9.
$$

Step 5: Use logarithm properties to combine the terms:

$$
\ln 8 - \ln 9 = \ln \left( \frac{8}{9} \right).
$$

Thus, the limit is

$$
\lim _{x \rightarrow 0} \frac{8^{x}-9^{x}}{x} = \ln \left( \frac{8}{9} \right).
$$
The limit is $$ \ln \left( \frac{8}{9} \right) $$.

Evaluate the limit using L’Hospital’s rule