Evaluate limit when x approaches to 5 and the function is (x^7-5^7)/(x-5)

To evaluate the limit \(\lim_{{x \to 5}} \frac{{x^7 - 5^7}}{{x - 5}}\), we can use L'Hôpital's Rule. This rule is applicable when the limit results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

Given the function:

\[
f(x) = \frac{{x^7 - 5^7}}{{x - 5}}
\]

we first substitute \(x = 5\):

\[
f(5) = \frac{{5^7 - 5^7}}{{5 - 5}} = \frac{0}{0}
\]

This is an indeterminate form, so we apply L'Hôpital's Rule, which states that:

\[
\lim_{{x \to c}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to c}} \frac{{f'(x)}}{{g'(x)}}
\]

if the limit on the right-hand side exists. Here, \(f(x) = x^7 - 5^7\) and \(g(x) = x - 5\).

First, we find the derivatives:

\[
f'(x) = \frac{d}{dx}(x^7 - 5^7) = 7x^6
\]

\[
g'(x) = \frac{d}{dx}(x - 5) = 1
\]

Now we apply L'Hôpital's Rule:

\[
\lim_{{x \to 5}} \frac{{x^7 - 5^7}}{{x - 5}} = \lim_{{x \to 5}} \frac{{7x^6}}{1} = 7 \cdot 5^6
\]

Calculating \(5^6\):

\[
5^6 = 15625
\]

Thus:

\[
7 \cdot 15625 = 109375
\]

Therefore, the limit is:

\[
\boxed{109375}
\]

Would you like more details or have any questions? Here are some related questions you might be interested in:

1. How do you evaluate limits using L'Hôpital's Rule?
2. What are the conditions for using L'Hôpital's Rule?
3. Can you provide an example of evaluating a limit without L'Hôpital's Rule?
4. How do you find the derivative of a polynomial function?
5. What are indeterminate forms in limits?

**Tip:** Always check the form of a limit before applying L'Hôpital's Rule to ensure it results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

Learn how to evaluate the limit \(\lim_{x \to 5} \frac{x^7 - 5^7}{x - 5}\) using L'Hôpital's Rule. Detailed step-by-step solution provided, including the use of derivatives and mathematical reasoning. Suitable for advanced high school students and anyone preparing for calculus examinations.

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