1. **State the problem:** Find the limit $$\lim_{x\to \infty}\frac{5^{x+1} + 7^{x+1}}{5^x - 7^x}$$ without using differentiation or L'Hopital's Rule. 2. **Rewrite the expression:** Use the property of exponents $$a^{x+1} = a \cdot a^x$$ to get $$\frac{5 \cdot 5^x + 7 \cdot 7^x}{5^x - 7^x}$$. 3. **Factor common terms:** Factor $$7^x$$ from numerator and denominator because as $$x\to \infty$$, $$7^x$$ grows faster than $$5^x$$. So, $$\frac{5 \cdot 5^x + 7 \cdot 7^x}{5^x - 7^x} = \frac{7^x \left(5 \cdot \left(\frac{5}{7}\right)^x + 7\right)}{7^x \left(\left(\frac{5}{7}\right)^x - 1\right)}$$. 4. **Simplify the fractions:** Cancel $$7^x$$ which is nonzero for large $$x$$, $$= \frac{5 \left(\frac{5}{7}\right)^x + 7}{\left(\frac{5}{7}\right)^x - 1}$$. 5. **Analyze the limit:** Since $$\frac{5}{7} < 1$$, we have $$\lim_{x \to \infty} \left(\frac{5}{7}\right)^x = 0.$$ 6. **Substitute the limit:** $$\lim_{x \to \infty} \frac{5 \cdot 0 + 7}{0 - 1} = \frac{7}{-1} = -7.$$