Question

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow 2} \frac{5^{x}-25}{x-2}$$

Short answer

Question: Evaluate the following limit exactly: $$\lim _{x \rightarrow 2} \frac{5^x-25}{x-2}$$ Answer: The value of the given limit exactly is $$25 \ln 5$$.

Step-by-step solution

Check if the limit is in indeterminate form

Plug in x = 2 into the given limit to see if the function is in indeterminate form: $$\frac{5^{2}-25}{2-2} = \frac{25 - 25}{0} = \frac{0}{0}$$ Since the function is in indeterminate form, we may proceed with L'Hôpital's Rule.

Apply L'Hôpital's Rule

Differentiate the numerator and the denominator with respect to x: Numerator: \(\frac{d}{dx}(5^x - 25) = (\ln 5)5^x\) Denominator: \(\frac{d}{dx}(x - 2) = 1\) Rewrite the limit using the derivatives of the numerator and the denominator: $$\lim _{x \rightarrow 2} \frac{(\ln 5)5^x}{1}$$

Evaluate the limit

Now plug in x = 2 into the limit: $$\lim _{x \rightarrow 2} (\ln 5)5^x=\left(\ln 5\right) 5^2$$ Hence, the value of the limit is: $$\boxed{25 \ln 5}$$