Select the correct alternatives out of given four alternatives in each.
`lim_(xto0)((1+x)^(16)-1)/((1+x)^(4)-1)` is-

Options

A. 0
B. 4
C. 8
D. 16

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{(1+x)^{16}-1}{(1+x)^{4}-1} \), we will follow these steps:

### Step 1: Direct Substitution
First, we will try to substitute \( x = 0 \) directly into the limit.

\[ \text{Substituting } x = 0: \quad (1+0)^{16} - 1 = 1^{16} - 1 = 0 \]
\[ (1+0)^{4} - 1 = 1^{4} - 1 = 0 \]

Since both the numerator and denominator evaluate to 0, we have an indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter \( \frac{0}{0} \) form, consider using L'Hospital's Rule.

### Step 2: Apply L'Hospital's Rule
Since we have the indeterminate form, we can apply L'Hospital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

**Numerator:**
\[ \text{Let } f(x) = (1+x)^{16} - 1 \implies f'(x) = 16(1+x)^{15} \]

**Denominator:**
\[ \text{Let } g(x) = (1+x)^{4} - 1 \implies g'(x) = 4(1+x)^{3} \]

### Step 3: Rewrite the Limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{16(1+x)^{15}}{4(1+x)^{3}} \]

### Step 4: Simplify the Expression
We can simplify the expression:

\[ = \lim_{x \to 0} \frac{16}{4} \cdot \frac{(1+x)^{15}}{(1+x)^{3}} = \lim_{x \to 0} 4(1+x)^{12} \]

### Step 5: Evaluate the Limit
Now we substitute \( x = 0 \):

\[ = 4(1+0)^{12} = 4 \cdot 1 = 4 \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to 0} \frac{(1+x)^{16}-1}{(1+x)^{4}-1} = 4 \]

The correct alternative is option 2, which is 4.

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