If `f(x) = {((1+(4x)/5)^(1/x),x!=0),(e^(4/5),x=0):}`, then

Options

A. `lim_(x to 0) f(x) = e^(2/5)`
B. `lim_(x to 0) f(x)` does not exist
C. f(x) is continuous at x = 0
D. f(x) is discontinuous at x = 0

Answer

To determine whether the function

\[ f(x) = \begin{cases} \left(1 + \frac{4x}{5}\right)^{\frac{1}{x}} & \text{if } x \neq 0 \\ e^{\frac{4}{5}} & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to check the following condition for continuity:

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

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

From the definition of the function, we have:

\[ f(0) = e^{\frac{4}{5}} \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

For \( x \neq 0 \), we need to evaluate:

\[ \lim_{x \to 0} \left(1 + \frac{4x}{5}\right)^{\frac{1}{x}} \]

This expression is of the form \( 1^{\infty} \) as \( x \to 0 \). To resolve this, we can use the exponential limit property:

\[ \lim_{x \to 0} a^b = e^{\lim_{x \to 0} b \ln(a)} \]

where \( a = 1 + \frac{4x}{5} \) and \( b = \frac{1}{x} \).

### Step 3: Rewrite the limit

We can rewrite the limit as follows:

\[ \lim_{x \to 0} \left(1 + \frac{4x}{5}\right)^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{1}{x} \ln\left(1 + \frac{4x}{5}\right)} \]

### Step 4: Evaluate \( \ln\left(1 + \frac{4x}{5}\right) \)

Using the Taylor expansion for \( \ln(1 + u) \) around \( u = 0 \):

\[ \ln(1 + u) \approx u \quad \text{for small } u \]

we have:

\[ \ln\left(1 + \frac{4x}{5}\right) \approx \frac{4x}{5} \]

### Step 5: Substitute back into the limit

Now substituting this back into our limit:

\[ \lim_{x \to 0} \frac{1}{x} \ln\left(1 + \frac{4x}{5}\right) \approx \lim_{x \to 0} \frac{1}{x} \cdot \frac{4x}{5} = \lim_{x \to 0} \frac{4}{5} = \frac{4}{5} \]

### Step 6: Final limit calculation

Thus, we have:

\[ \lim_{x \to 0} f(x) = e^{\frac{4}{5}} \]

### Step 7: Compare limits

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

\[ \lim_{x \to 0} f(x) = e^{\frac{4}{5}} = f(0) \]

### Conclusion

Since

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

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

---