`"If "y=f(x^(3)),z=g(x^(5)),f'(x)=tan x, and g'(x) = sec x, " then find the value of of "lim_(xrarr0) ((dy//dz))/(x)`

Explanation

To solve the problem, we need to find the limit:

\[ \lim_{x \to 0} \frac{dy/dz}{x} \]

where \( y = f(x^3) \) and \( z = g(x^5) \) with \( f'(x) = \tan x \) and \( g'(x) = \sec x \).

### Step-by-step Solution:

1. **Differentiate \( y \) with respect to \( x \)**:
\[ y = f(x^3) \]
Using the chain rule:
\[ \frac{dy}{dx} = f'(x^3) \cdot \frac{d}{dx}(x^3) = f'(x^3) \cdot 3x^2 \]

2. **Differentiate \( z \) with respect to \( x \)**:
\[ z = g(x^5) \]
Again, using the chain rule:
\[ \frac{dz}{dx} = g'(x^5) \cdot \frac{d}{dx}(x^5) = g'(x^5) \cdot 5x^4 \]

3. **Find \( \frac{dy}{dz} \)**:
Using the formula:
\[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \]
Substitute the derivatives found in the previous steps:
\[ \frac{dy}{dz} = \frac{f'(x^3) \cdot 3x^2}{g'(x^5) \cdot 5x^4} \]

4. **Simplify \( \frac{dy}{dz} \)**:
\[ \frac{dy}{dz} = \frac{3f'(x^3)}{5g'(x^5)} \cdot \frac{x^2}{x^4} = \frac{3f'(x^3)}{5g'(x^5)} \cdot \frac{1}{x^2} \]

5. **Substitute \( f'(x) \) and \( g'(x) \)**:
Given \( f'(x) = \tan x \) and \( g'(x) = \sec x \):
\[ f'(x^3) = \tan(x^3) \quad \text{and} \quad g'(x^5) = \sec(x^5) \]
Thus,
\[ \frac{dy}{dz} = \frac{3 \tan(x^3)}{5 \sec(x^5)} \cdot \frac{1}{x^2} \]

6. **Find the limit**:
Now we need to find:
\[ \lim_{x \to 0} \frac{dy/dz}{x} = \lim_{x \to 0} \frac{3 \tan(x^3)}{5 \sec(x^5) \cdot x^2} \cdot \frac{1}{x} \]
Simplifying this gives:
\[ = \lim_{x \to 0} \frac{3 \tan(x^3)}{5 \sec(x^5) \cdot x^3} \]

7. **Evaluate the limit**:
As \( x \to 0 \), \( \tan(x^3) \approx x^3 \) and \( \sec(x^5) \approx 1 \):
\[ \lim_{x \to 0} \frac{3 \tan(x^3)}{5 \sec(x^5) \cdot x^3} = \lim_{x \to 0} \frac{3x^3}{5 \cdot 1 \cdot x^3} = \frac{3}{5} \]

### Final Answer:
\[ \lim_{x \to 0} \frac{dy/dz}{x} = \frac{3}{5} \]