To find the limit of the given function as \(x\) approaches 0, we need to evaluate \(\lim\limits_{x \to 0}x^3(\text{cosec} \ x)^2\).

The function involves trigonometric functions. Recall that \(\text{cosec} \ x = \frac{1}{\sin x}\). So, we can rewrite the expression as:

$$ \lim\limits_{x \to 0} x^3 \left(\frac{1}{\sin x}\right)^2 = \lim\limits_{x \to 0} \frac{x^3}{\sin^2 x} $$

If we directly substitute \(x=0\) into the expression, we get \(\frac{0^3}{\sin^2 0} = \frac{0}{0}\), which is an indeterminate form. This indicates that we need to use other methods, such as standard limits or L'Hopital's rule.

Understanding the Limit Problem

The problem asks for the limit of the function \(f(x) = x^3(\text{cosec} \ x)^2\) as \(x\) approaches 0. This is a common type of limit problem involving trigonometric functions near 0.

The key standard limit involving \(\sin x\) near 0 is \(\lim\limits_{x \to 0} \frac{\sin x}{x} = 1\). We can manipulate the expression to use this standard limit.

Step-by-Step Limit Calculation

Let's rewrite the expression \(\frac{x^3}{\sin^2 x}\) to utilize the standard limit \(\lim\limits_{x \to 0} \frac{\sin x}{x} = 1\).

We have \(\frac{x^3}{\sin^2 x} = \frac{x^3}{(\sin x)^2}\). We can separate the terms to get something like \(\frac{x}{\sin x}\) or \(\frac{\sin x}{x}\).

$$ \frac{x^3}{\sin^2 x} = \frac{x \cdot x^2}{\sin^2 x} = x \cdot \frac{x^2}{\sin^2 x} = x \cdot \left(\frac{x}{\sin x}\right)^2 $$

Now, we can take the limit as \(x \to 0\):

$$ \lim\limits_{x \to 0} x \cdot \left(\frac{x}{\sin x}\right)^2 $$

Using the property that the limit of a product is the product of the limits (if the individual limits exist), we get:

$$ \left(\lim\limits_{x \to 0} x\right) \cdot \left(\lim\limits_{x \to 0} \frac{x}{\sin x}\right)^2 $$

We evaluate each limit separately:

• The first limit is straightforward: \(\lim\limits_{x \to 0} x = 0\).
• The second limit uses the standard result \(\lim\limits_{x \to 0} \frac{\sin x}{x} = 1\). Therefore, \(\lim\limits_{x \to 0} \frac{x}{\sin x} = \frac{1}{\lim\limits_{x \to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1\).

Substitute these values back into the expression:

$$ (0) \cdot (1)^2 = 0 \cdot 1 = 0 $$

Thus, the limit \(\lim\limits_{x \to 0}x^3(\text{cosec} \ x)^2\) is equal to 0.

Verification of the Limit Result

The calculation shows that the limit is 0. This corresponds to one of the given options.

Revision Table: Key Limit Concepts

Additional Information: Trigonometric Limits

Trigonometric limits are fundamental in calculus. Understanding the behavior of trigonometric functions near certain points, especially 0, is crucial for evaluating limits involving these functions. The standard limit \(\lim\limits_{x \to 0} \frac{\sin x}{x} = 1\) is derived using the Squeeze Theorem and forms the basis for evaluating many other trigonometric limits, including \(\lim\limits_{x \to 0} \frac{\tan x}{x} = 1\) and \(\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\).

In this problem, rewriting \(\text{cosec} \ x\) in terms of \(\sin x\) was the first step to simplifying the expression and applying standard limit techniques.

Using L'Hopital's rule is another valid method for indeterminate forms, but sometimes algebraic manipulation using standard limits is faster and simpler, as seen in this case.