`lim_(x to oo) (1+k/x)^(mx)` is equal to

Options

A. `m^k`
B. `e^m`
C. `e^(mk)`
D. `e^(m//k)`

Answer

To solve the limit \( \lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^{mx} \), we can follow these steps:

### Step 1: Identify the form of the limit
As \( x \to \infty \), \( \frac{k}{x} \to 0 \). Thus, \( 1 + \frac{k}{x} \to 1 \). The expression becomes \( 1^{\infty} \), which is an indeterminate form.

**Hint:** Recognize that \( 1^{\infty} \) is an indeterminate form that requires further manipulation.

### Step 2: Rewrite the limit using the exponential function
We can rewrite the expression using the exponential function:
\[ \lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^{mx} = \lim_{x \to \infty} e^{mx \ln\left(1 + \frac{k}{x}\right)} \]

**Hint:** Use the property \( a^b = e^{b \ln a} \) to express the limit in terms of the exponential function.

### Step 3: Simplify \( \ln\left(1 + \frac{k}{x}\right) \)
Using the Taylor series expansion for \( \ln(1 + u) \) where \( u = \frac{k}{x} \):
\[ \ln\left(1 + \frac{k}{x}\right) \approx \frac{k}{x} \quad \text{as } x \to \infty \]

**Hint:** Recall that for small \( u \), \( \ln(1 + u) \approx u \).

### Step 4: Substitute back into the limit
Substituting this approximation back into our limit gives:
\[ mx \ln\left(1 + \frac{k}{x}\right) \approx mx \cdot \frac{k}{x} = mk \]

**Hint:** Notice how the \( x \) in the numerator and denominator cancels out.

### Step 5: Evaluate the limit
Now we can evaluate the limit:
\[ \lim_{x \to \infty} e^{mx \ln\left(1 + \frac{k}{x}\right)} = e^{mk} \]

**Hint:** The limit of the exponential function simplifies to just the exponent.

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^{mx} = e^{mk} \]