`"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 1: Differentiate \( y \) with respect to \( x \)

Using the chain rule, we have:

\[ \frac{dy}{dx} = f'(x^3) \cdot \frac{d}{dx}(x^3) = f'(x^3) \cdot 3x^2 \]

Substituting \( f'(x) = \tan x \):

\[ \frac{dy}{dx} = \tan(x^3) \cdot 3x^2 \]

### Step 2: Differentiate \( z \) with respect to \( x \)

Similarly, for \( z \):

\[ \frac{dz}{dx} = g'(x^5) \cdot \frac{d}{dx}(x^5) = g'(x^5) \cdot 5x^4 \]

Substituting \( g'(x) = \sec x \):

\[ \frac{dz}{dx} = \sec(x^5) \cdot 5x^4 \]

### Step 3: Find \( \frac{dy}{dz} \)

Using the formula for derivatives:

\[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{\tan(x^3) \cdot 3x^2}{\sec(x^5) \cdot 5x^4} \]

This simplifies to:

\[ \frac{dy}{dz} = \frac{3 \tan(x^3)}{5 \sec(x^5) \cdot x^2} \]

### Step 4: Substitute into the limit

Now we substitute this into the limit:

\[ \lim_{x \to 0} \frac{dy/dz}{x} = \lim_{x \to 0} \frac{3 \tan(x^3)}{5 \sec(x^5) \cdot x^3} \]

### Step 5: Evaluate the limit

As \( x \to 0 \):

- \( \tan(x^3) \approx x^3 \) (since \( \tan u \approx u \) for small \( u \))
- \( \sec(x^5) \approx 1 \) (since \( \sec u \approx 1 \) for small \( u \))

Thus, we have:

\[ \lim_{x \to 0} \frac{3 \tan(x^3)}{5 \sec(x^5) \cdot x^3} = \lim_{x \to 0} \frac{3 \cdot x^3}{5 \cdot 1 \cdot x^3} = \frac{3}{5} \]

### Final Answer

The value of the limit is:

\[ \frac{3}{5} \]