To evaluate the limit

$\lim_{x \to \infty} (10x^2 + 5)^{\frac{6}{x}}$

we'll analyze the behavior of $(10x^2 + 5)^{\frac{6}{x}}$ as $x \to \infty$.

1. Rewrite the expression by taking the natural logarithm to make it easier to evaluate: $y = (10x^2 + 5)^{\frac{6}{x}}$ Taking the natural logarithm of both sides gives: $\ln y = \frac{6}{x} \ln(10x^2 + 5)$

2. Now, we need to find the limit of $\ln y$ as $x \to \infty$: $\lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{6 \ln(10x^2 + 5)}{x}$

3. We can use L'Hôpital's Rule here since the form is $\frac{\infty}{\infty}$: $\lim_{x \to \infty} \frac{6 \ln(10x^2 + 5)}{x} = \lim_{x \to \infty} \frac{6 \cdot \frac{1}{10x^2 + 5} \cdot 20x}{1} = \lim_{x \to \infty} \frac{120x}{10x^2 + 5}$

4. Simplify the fraction: $\lim_{x \to \infty} \frac{120x}{10x^2 + 5} = \lim_{x \to \infty} \frac{120}{10x + \frac{5}{x}}$

5. As $x \to \infty$, the term $\frac{5}{x} \to 0$, so we get: $\lim_{x \to \infty} \frac{120}{10x} = \lim_{x \to \infty} \frac{12}{x} = 0$

6. Therefore, $\lim_{x \to \infty} \ln y = 0$, which implies that $\lim_{x \to \infty} y = e^0 = 1$.

Thus, the answer is:

$\lim_{x \to \infty} (10x^2 + 5)^{\frac{6}{x}} = 1$

Would you like further details on this solution, or do you have any questions?

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Here are some related questions to explore:

1. What is the limit $\lim_{x \to \infty} (ax^n + b)^{\frac{c}{x}}$ for different values of $a$, $b$, $c$, and $n$?
2. How does the natural logarithm help simplify limits involving exponentials?
3. Can you explain why L'Hôpital's Rule is useful for evaluating indeterminate forms?
4. What happens if the exponent $\frac{6}{x}$ is replaced by $\frac{k}{x}$ with $k$ as a variable constant?
5. How would you evaluate the limit if the exponent was $\frac{6}{\sqrt{x}}$ instead?

Tip: When dealing with limits at infinity involving exponentials, taking logarithms can simplify the problem, especially when the exponent approaches zero.