Show that the function `f(x)={{:(,x^2,x le 1),(,1/x,x gt 1):}` continuous at x=1 but not differentiable.

Answer

To show that the function \( f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ \frac{1}{x} & \text{if } x > 1 \end{cases} \) is continuous at \( x = 1 \) but not differentiable, we will follow these steps:

### Step 1: Check Continuity at \( x = 1 \)

To check if \( f(x) \) is continuous at \( x = 1 \), we need to show that:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

#### Step 1.1: Calculate \( f(1) \)

Since \( 1 \leq 1 \), we use the first case of the function:

\[ f(1) = 1^2 = 1 \]

#### Step 1.2: Calculate the Left-Hand Limit (LHL)

We calculate the limit as \( x \) approaches 1 from the left:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1^2 = 1 \]

#### Step 1.3: Calculate the Right-Hand Limit (RHL)

Now we calculate the limit as \( x \) approaches 1 from the right:

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = \frac{1}{1} = 1 \]

#### Step 1.4: Conclusion on Continuity

Since:

\[ \lim_{x \to 1^-} f(x) = 1, \quad \lim_{x \to 1^+} f(x) = 1, \quad \text{and } f(1) = 1 \]

We conclude that:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

Thus, \( f(x) \) is continuous at \( x = 1 \).

### Step 2: Check Differentiability at \( x = 1 \)

To show that \( f(x) \) is not differentiable at \( x = 1 \), we need to calculate the left-hand derivative (LHD) and the right-hand derivative (RHD) at that point.

#### Step 2.1: Calculate the Left-Hand Derivative (LHD)

The left-hand derivative is given by:

\[ \text{LHD} = \lim_{h \to 0} \frac{f(1 - h) - f(1)}{(1 - h) - 1} \]

Substituting \( f(1 - h) \):

\[ \text{LHD} = \lim_{h \to 0} \frac{(1 - h)^2 - 1}{-h} \]

Calculating \( (1 - h)^2 - 1 \):

\[ (1 - h)^2 - 1 = 1 - 2h + h^2 - 1 = -2h + h^2 \]

So we have:

\[ \text{LHD} = \lim_{h \to 0} \frac{-2h + h^2}{-h} = \lim_{h \to 0} \frac{h(2 - h)}{h} = \lim_{h \to 0} (2 - h) = 2 \]

#### Step 2.2: Calculate the Right-Hand Derivative (RHD)

The right-hand derivative is given by:

\[ \text{RHD} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{(1 + h) - 1} \]

Substituting \( f(1 + h) \):

\[ \text{RHD} = \lim_{h \to 0} \frac{\frac{1}{1 + h} - 1}{h} \]

Calculating \( \frac{1}{1 + h} - 1 \):

\[ \frac{1}{1 + h} - 1 = \frac{1 - (1 + h)}{1 + h} = \frac{-h}{1 + h} \]

So we have:

\[ \text{RHD} = \lim_{h \to 0} \frac{-h/(1 + h)}{h} = \lim_{h \to 0} \frac{-1}{1 + h} = -1 \]

#### Step 2.3: Conclusion on Differentiability

Since:

\[ \text{LHD} = 2 \quad \text{and} \quad \text{RHD} = -1 \]

We conclude that the left-hand derivative is not equal to the right-hand derivative:

\[ \text{LHD} \neq \text{RHD} \]

Thus, \( f(x) \) is not differentiable at \( x = 1 \).

### Final Conclusion

The function \( f(x) \) is continuous at \( x = 1 \) but not differentiable at that point.

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