\(\lim _{\mathrm{h} \rightarrow 5}\left[\left\\{(2 \mathrm{~h}+5)^{(5 / 2)}-(15)^{(5 / 2)}\right\\} /\left(\mathrm{h}^{3}-125\right)\right]=?\) (a) \(\sqrt{5}\) (b) \(\sqrt{125}\) (c) \(\sqrt{15}\) (d) \((15)^{(5 / 2)}\)

First, also recall that we typically use the notation \(\lim_{x \rightarrow a}\) to denote the limit as x approaches a value a. In our case, we'd have the given limit as \(\lim_{h \rightarrow 5}\). Now, rewrite the expression in terms of the given limit: \[ \lim_{h \rightarrow 5} \frac{(2h +5)^{\frac{5}{2}} - (15)^{\frac{5}{2}}}{h^3 - 125} \]

Next, we need to identify the indeterminate form of this limit by substituting the given value of h, which is 5: \[ \frac{(2(5) + 5)^{\frac{5}{2}} - (15)^{\frac{5}{2}}}{(5)^3 - 125} \] We can simplify this as: \[ \frac{(15)^{\frac{5}{2}} - (15)^{\frac{5}{2}}}{(125) - 125} \] Which looks like: \[ \frac{0}{0} \] This is an indeterminate form.

Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule for finding the limit. L'Hôpital's Rule states that if \(\lim_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), then the limit can be found by differentiating both the numerator and denominator: \[ \lim_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \lim_{x \rightarrow a} \dfrac{f'(x)}{g'(x)} \] Let's find the derivatives of the numerator and denominator: \(f(h) = (2h + 5)^{\frac{5}{2}} - (15)^{\frac{5}{2}}\) \(f'(h) = \frac{5}{2}(2h + 5)^{\frac{3}{2}}(2)\) \(g(h) = h^3 - 125\) \(g'(h) = 3h^2\) Now, using L'Hôpital's Rule, we plug these derived functions back in to find the limit: \[ \lim_{h \rightarrow 5} \frac{f'(h)}{g'(h)} = \lim_{h \rightarrow 5} \frac{\frac{5}{2}(2h + 5)^{\frac{3}{2}}(2)}{3h^2} \]

Finally, plug in the value h = 5: \[ \lim_{h \rightarrow 5} \frac{\frac{5}{2}(2(5) + 5)^{\frac{3}{2}}(2)}{3(5)^2} = \frac{\frac{5}{2}(15)^{\frac{3}{2}}(2)}{3(25)} \] Simplify the expression: \[ \frac{5(15)^{\frac{3}{2}}}{3(25)} \] This reduces to: \[ \frac{15^{\frac{5}{2}}}{\sqrt{15}} \] Thus, the answer is \(\boxed{\text{(c)} \sqrt{15}}\).