Math Snap

STEP 1

Assumptions
1. We need to find the limit: $$\lim _{x \rightarrow 0^{+}}(\tan (8 x))^{x}$$ 2. The function is in the indeterminate form $0^0$ as $x \to 0^+$.
3. We will use logarithmic properties and L'Hospital's Rule to solve this limit.

STEP 2

To handle the indeterminate form $0^0$, take the natural logarithm of the function. Let $y = (\tan(8x))^x$, then:
$$\ln y = x \ln(\tan(8x))$$

STEP 3

Now, find the limit of $\ln y$ as $x \to 0^+$:
$$\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} x \ln(\tan(8x))$$

STEP 4

This limit is in the form $0 \cdot (-\infty)$, which is indeterminate. Rewrite it as a fraction:
$$\lim_{x \to 0^+} x \ln(\tan(8x)) = \lim_{x \to 0^+} \frac{\ln(\tan(8x))}{1/x}$$

STEP 5

Now, the limit is in the form $\frac{-\infty}{\infty}$, which is suitable for L'Hospital's Rule. Differentiate the numerator and the denominator:
Numerator: $\frac{d}{dx}[\ln(\tan(8x))] = \frac{1}{\tan(8x)} \cdot \sec^2(8x) \cdot 8 = \frac{8 \sec^2(8x)}{\tan(8x)}$
Denominator: $\frac{d}{dx}[1/x] = -\frac{1}{x^2}$

STEP 6

Apply L'Hospital's Rule:
$$\lim_{x \to 0^+} \frac{\ln(\tan(8x))}{1/x} = \lim_{x \to 0^+} \frac{8 \sec^2(8x)}{\tan(8x)} \cdot (-x^2)$$

STEP 7

Simplify the expression:
$$= \lim_{x \to 0^+} -8x^2 \cdot \frac{\sec^2(8x)}{\tan(8x)}$$

STEP 8

As $x \to 0^+$, $\tan(8x) \approx 8x$ and $\sec^2(8x) \approx 1$. Substitute these approximations:
$$= \lim_{x \to 0^+} -8x^2 \cdot \frac{1}{8x}$$

STEP 9

Simplify further:
$$= \lim_{x \to 0^+} -x$$

STEP 10

Evaluate the limit:
$$= 0$$

STEP 11

Since $\lim_{x \to 0^+} \ln y = 0$, we have $\ln y = 0$, which implies $y = e^0 = 1$.

Thus, the original limit is:
$$\lim_{x \to 0^+} (\tan(8x))^x = 1$$