`lim_(x to 0) ((1 + x)^(5) -1)/((1 + x)^(3) - 1)` is equal to

Options

A. 6
B. `(9)/(4)`
C. `(5)/(3)`
D. `(3)/(5)`

Answer

To solve the limit \( \lim_{x \to 0} \frac{(1 + x)^5 - 1}{(1 + x)^3 - 1} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{(1 + x)^5 - 1}{(1 + x)^3 - 1} \]

### Step 2: Apply the binomial expansion
Using the binomial theorem, we can expand \( (1 + x)^5 \) and \( (1 + x)^3 \):
\[ (1 + x)^5 = 1 + 5x + \frac{5 \cdot 4}{2}x^2 + \frac{5 \cdot 4 \cdot 3}{6}x^3 + \ldots = 1 + 5x + 10x^2 + \frac{10}{3}x^3 + O(x^4) \]
\[ (1 + x)^3 = 1 + 3x + \frac{3 \cdot 2}{2}x^2 + \ldots = 1 + 3x + 3x^2 + O(x^3) \]

### Step 3: Substitute the expansions into the limit
Substituting these expansions into the limit gives:
\[ \lim_{x \to 0} \frac{(1 + 5x + 10x^2 + O(x^3)) - 1}{(1 + 3x + 3x^2 + O(x^3)) - 1} \]
This simplifies to:
\[ \lim_{x \to 0} \frac{5x + 10x^2 + O(x^3)}{3x + 3x^2 + O(x^3)} \]

### Step 4: Factor out \( x \)
Now, we can factor \( x \) out of both the numerator and the denominator:
\[ \lim_{x \to 0} \frac{x(5 + 10x + O(x^2))}{x(3 + 3x + O(x^2))} \]
This simplifies to:
\[ \lim_{x \to 0} \frac{5 + 10x + O(x^2)}{3 + 3x + O(x^2)} \]

### Step 5: Evaluate the limit as \( x \to 0 \)
As \( x \) approaches 0, the higher order terms \( O(x^2) \) vanish:
\[ \frac{5 + 0}{3 + 0} = \frac{5}{3} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{(1 + x)^5 - 1}{(1 + x)^3 - 1} = \frac{5}{3} \]

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