4. Evaluate the following limit (i) \( \lim _{x \rightarrow 0^{+}}(\csc x-\cot x) \) (ii) \( \lim _{x \rightarrow 0^{+}}\left(\frac{\sin 2 x}{\sqrt{5 x}}\right) \).

Let's evaluate each limit step by step. --- ### **Problem (i)** \[ \lim_{x \rightarrow 0^{+}} (\csc x - \cot x) \] **Step 1:** Express \(\csc x\) and \(\cot x\) in terms of sine and cosine: \[ \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x} \] So, \[ \csc x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x} \] **Step 2:** Analyze the limit as \(x \rightarrow 0^{+}\): \[ \lim_{x \rightarrow 0^{+}} \frac{1 - \cos x}{\sin x} \] As \(x \rightarrow 0\), \[ 1 - \cos x \approx \frac{x^2}{2}, \quad \sin x \approx x \] Thus, \[ \frac{1 - \cos x}{\sin x} \approx \frac{\frac{x^2}{2}}{x} = \frac{x}{2} \rightarrow 0 \] Alternatively, using **L'Hôpital's Rule** since the limit is of the form \(\frac{0}{0}\): \[ \lim_{x \rightarrow 0^{+}} \frac{1 - \cos x}{\sin x} = \lim_{x \rightarrow 0^{+}} \frac{\sin x}{\cos x} = \frac{0}{1} = 0 \] **Conclusion:** \[ \lim_{x \rightarrow 0^{+}} (\csc x - \cot x) = 0 \] --- ### **Problem (ii)** \[ \lim_{x \rightarrow 0^{+}} \left( \frac{\sin 2x}{\sqrt{5x}} \right) \] **Step 1:** Simplify the expression: \[ \frac{\sin 2x}{\sqrt{5x}} = \frac{2x}{\sqrt{5x}} \cdot \frac{\sin 2x}{2x} \] As \(x \rightarrow 0\), \[ \frac{\sin 2x}{2x} \rightarrow 1 \] So, \[ \frac{\sin 2x}{\sqrt{5x}} \approx \frac{2x}{\sqrt{5x}} = \frac{2}{\sqrt{5}} \cdot \sqrt{x} \] **Step 2:** Evaluate the limit: \[ \lim_{x \rightarrow 0^{+}} \frac{2}{\sqrt{5}} \cdot \sqrt{x} = \frac{2}{\sqrt{5}} \cdot 0 = 0 \] **Conclusion:** \[ \lim_{x \rightarrow 0^{+}} \left( \frac{\sin 2x}{\sqrt{5x}} \right) = 0 \] --- **Final Answer:** Both limits are equal to zero. \[ \boxed{0} \]