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

Question

UpStudy AI · Accepted Answer

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:**
$$
\lim_{x \rightarrow 0^{+}} \left( \frac{\sin 2x}{\sqrt{5x}} \right) = 0
$$

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**Final Answer:**

Both limits are equal to zero.

$$
\boxed{0}
$$
Both limits are 0. $$ \boxed{0} $$

Evaluate the following limit
(i)
(ii) .

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Answer

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Problem (i)

Problem (ii)

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Evaluate the following limit (i) (ii) .