Ascertain the existence of the following limits :
`lim_(xto0)(x^(2)sin(1//x))/(sinx )`

Answer

To ascertain the existence of the limit

\[ \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x}, \]

we can follow these steps:

### Step 1: Rewrite the Limit

We start by rewriting the limit in a more manageable form. We can express it as:

\[ \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x} = \lim_{x \to 0} \left( \frac{x^2}{\sin x} \cdot \sin\left(\frac{1}{x}\right) \right). \]

### Step 2: Analyze Each Part of the Limit

We know from standard limits that:

\[ \lim_{x \to 0} \frac{x}{\sin x} = 1. \]

Thus,

\[ \lim_{x \to 0} \frac{x^2}{\sin x} = \lim_{x \to 0} x \cdot \frac{x}{\sin x} = 0 \cdot 1 = 0. \]

### Step 3: Evaluate the Second Part

Next, we need to evaluate the limit of \(\sin\left(\frac{1}{x}\right)\) as \(x\) approaches 0. As \(x\) approaches 0, \(\frac{1}{x}\) approaches infinity. The sine function oscillates between -1 and 1 for all real numbers, including infinity. Therefore, we can say:

\[ \sin\left(\frac{1}{x}\right) \text{ is bounded between } -1 \text{ and } 1. \]

### Step 4: Combine the Results

Now we combine our findings:

\[ \lim_{x \to 0} \left( \frac{x^2}{\sin x} \cdot \sin\left(\frac{1}{x}\right) \right) = \lim_{x \to 0} \frac{x^2}{\sin x} \cdot \lim_{x \to 0} \sin\left(\frac{1}{x}\right). \]

Since \(\lim_{x \to 0} \frac{x^2}{\sin x} = 0\) and \(\sin\left(\frac{1}{x}\right)\) is bounded, we can conclude:

\[ \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x} = 0 \cdot \text{(bounded value)} = 0. \]

### Final Result

Thus, the limit exists and is equal to:

\[ \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x} = 0. \]

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