Examine the continuity of the following function at the indicated pionts.
`f(x)={{:(,x^(2) cos (1/x) , x ne 0),(,0 , x =0):}" at x=0"`

Answer

To examine the continuity of the function

\[ f(x) = \begin{cases} x^2 \cos\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

at \(x = 0\), we need to check the following conditions:

1. \(f(0)\) is defined.
2. The left-hand limit as \(x\) approaches 0 exists.
3. The right-hand limit as \(x\) approaches 0 exists.
4. The left-hand limit, right-hand limit, and \(f(0)\) are all equal.

### Step 1: Find \(f(0)\)

Since \(f(x) = 0\) when \(x = 0\), we have:

\[ f(0) = 0 \]

### Step 2: Find the left-hand limit as \(x\) approaches 0

The left-hand limit is given by:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 \cos\left(\frac{1}{x}\right) \]

To evaluate this limit, we can substitute \(x = -h\) where \(h \to 0^+\):

\[ \lim_{h \to 0^+} (-h)^2 \cos\left(\frac{1}{-h}\right) = \lim_{h \to 0^+} h^2 \cos\left(-\frac{1}{h}\right) = \lim_{h \to 0^+} h^2 \cos\left(\frac{1}{h}\right) \]

### Step 3: Evaluate the limit

As \(h\) approaches 0, \(h^2\) approaches 0. The cosine function oscillates between -1 and 1, so we can say:

\[ -h^2 \leq h^2 \cos\left(\frac{1}{h}\right) \leq h^2 \]

By the Squeeze Theorem:

\[ \lim_{h \to 0^+} h^2 \cos\left(\frac{1}{h}\right) = 0 \]

Thus, the left-hand limit is:

\[ \lim_{x \to 0^-} f(x) = 0 \]

### Step 4: Find the right-hand limit as \(x\) approaches 0

The right-hand limit is given by:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 \cos\left(\frac{1}{x}\right) \]

Using a similar argument as above, we can substitute \(x = h\) where \(h \to 0^+\):

\[ \lim_{h \to 0^+} h^2 \cos\left(\frac{1}{h}\right) \]

Again, by the Squeeze Theorem, we find:

\[ \lim_{h \to 0^+} h^2 \cos\left(\frac{1}{h}\right) = 0 \]

Thus, the right-hand limit is:

\[ \lim_{x \to 0^+} f(x) = 0 \]

### Step 5: Check if the limits and function value are equal

Now we have:

- \(f(0) = 0\)
- \(\lim_{x \to 0^-} f(x) = 0\)
- \(\lim_{x \to 0^+} f(x) = 0\)

Since all three values are equal, we conclude that:

\[ f(x) \text{ is continuous at } x = 0. \]

### Final Conclusion

The function \(f(x)\) is continuous at \(x = 0\).

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