If `f(x)={{:((e^([2x]+2x+1)-1)/([2x]+2x+1),:,x ne 0),(1,":", x =0):}`, then (where `[.]` represents the greatest integer function)

Options

A. `lim_(xrarr0^(+))f(x)=1`
B. `lim_(xrarr0^(-))f(x)=e-1`
C. `f(x)` is continuous at x = 0
D. `f(x)` is discontinuous at x = 0

Answer

To determine the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \frac{e^{(2x + 2x + 1)} - 1}{(2x + 2x + 1)} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

we need to check the limit of \( f(x) \) as \( x \) approaches 0 from both sides and compare it to \( f(0) \).

### Step 1: Calculate the limit as \( x \to 0^+ \)

For \( x \to 0^+ \):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{(2x + 2x + 1)} - 1}{(2x + 2x + 1)} = \lim_{x \to 0^+} \frac{e^{(4x + 1)} - 1}{(4x + 1)} \]

As \( x \to 0^+ \), \( 4x + 1 \to 1 \):

\[ = \frac{e^1 - 1}{1} = e - 1 \]

### Step 2: Calculate the limit as \( x \to 0^- \)

For \( x \to 0^- \):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{e^{(2x + 2x + 1)} - 1}{(2x + 2x + 1)} = \lim_{x \to 0^-} \frac{e^{(4x + 1)} - 1}{(4x + 1)} \]

Again, as \( x \to 0^- \), \( 4x + 1 \to 1 \):

\[ = \frac{e^1 - 1}{1} = e - 1 \]

### Step 3: Compare limits with \( f(0) \)

Now we compare the limits with \( f(0) \):

\[ f(0) = 1 \]

### Step 4: Check continuity

For \( f(x) \) to be continuous at \( x = 0 \), we need:

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

Since:

\[ \lim_{x \to 0^+} f(x) = e - 1 \quad \text{and} \quad \lim_{x \to 0^-} f(x) = e - 1 \]

We find:

\[ \lim_{x \to 0} f(x) = e - 1 \neq f(0) = 1 \]

Thus, \( f(x) \) is discontinuous at \( x = 0 \).

### Conclusion

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

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