Evaluate `lim_(xtooo) ((7x^(2)+1)/(5x^(2)-1))^((x^(5))/(1-x^(3))).`

Explanation

To evaluate the limit

\[ \lim_{x \to \infty} \left(\frac{7x^2 + 1}{5x^2 - 1}\right)^{\frac{x^5}{1 - x^3}}, \]

we will follow these steps:

### Step 1: Simplify the base of the limit
We first simplify the expression inside the limit as \(x\) approaches infinity.

\[ \frac{7x^2 + 1}{5x^2 - 1} \]

Factoring out \(x^2\) from the numerator and denominator gives:

\[ \frac{7 + \frac{1}{x^2}}{5 - \frac{1}{x^2}}. \]

Now, taking the limit as \(x\) approaches infinity:

\[ \lim_{x \to \infty} \frac{7 + \frac{1}{x^2}}{5 - \frac{1}{x^2}} = \frac{7 + 0}{5 - 0} = \frac{7}{5}. \]

### Step 2: Simplify the exponent of the limit
Next, we simplify the exponent:

\[ \frac{x^5}{1 - x^3}. \]

Factoring out \(x^3\) from the denominator gives:

\[ \frac{x^5}{1 - x^3} = \frac{x^5}{-x^3(1 - \frac{1}{x^3})} = -\frac{x^2}{1 - \frac{1}{x^3}}. \]

Taking the limit as \(x\) approaches infinity:

\[ \lim_{x \to \infty} -\frac{x^2}{1 - \frac{1}{x^3}} = -\infty. \]

### Step 3: Combine the results
Now we combine the results from Steps 1 and 2:

\[ \lim_{x \to \infty} \left(\frac{7}{5}\right)^{-\infty}. \]

Since \(\frac{7}{5} > 1\), raising it to the power of \(-\infty\) results in:

\[ \left(\frac{7}{5}\right)^{-\infty} = 0. \]

### Final Result
Thus, the limit is:

\[ \lim_{x \to \infty} \left(\frac{7x^2 + 1}{5x^2 - 1}\right)^{\frac{x^5}{1 - x^3}} = 0. \]