Use l’Hôpital’s rule to find the limits in Exercises 7–52.
37. lim (y → 0) (√(5y + 25) - 5) / y

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First, identify the form of the limit as \( y \to 0 \) for the expression \( \frac{\sqrt{5y + 25} - 5}{y} \). Substitute \( y = 0 \) to check if it results in an indeterminate form.

Since substituting \( y = 0 \) gives \( \frac{\sqrt{25} - 5}{0} = \frac{5 - 5}{0} = \frac{0}{0} \), which is an indeterminate form, we can apply l'Hôpital's Rule.

Apply l'Hôpital's Rule by differentiating the numerator and denominator separately with respect to \( y \). The numerator is \( \sqrt{5y + 25} - 5 \), so find its derivative:

The derivative of the numerator is \( \frac{d}{dy} \left( \sqrt{5y + 25} - 5 \right) = \frac{1}{2\sqrt{5y + 25}} \times 5 = \frac{5}{2\sqrt{5y + 25}} \). The derivative of the denominator \( y \) is 1.

Rewrite the limit using these derivatives: \( \lim_{y \to 0} \frac{5}{2\sqrt{5y + 25}} \). Then, evaluate this new limit by substituting \( y = 0 \).

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