Evaluate the following limits .
`Lim_(x to 0) ((1+x)^(6)-1)/((1+x)^(2)-1)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1}, \]

we start by substituting \( x = 0 \):

1. **Substituting \( x = 0 \)**:
\[ \frac{(1+0)^6 - 1}{(1+0)^2 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}. \]
This gives us an indeterminate form \( \frac{0}{0} \).

**Hint**: When you encounter \( \frac{0}{0} \), consider using L'Hôpital's Rule, which allows you to differentiate the numerator and denominator.

2. **Applying L'Hôpital's Rule**:
We differentiate the numerator and the denominator separately.

- **Numerator**: Differentiate \( (1+x)^6 - 1 \):
\[ \frac{d}{dx}((1+x)^6) = 6(1+x)^5. \]

- **Denominator**: Differentiate \( (1+x)^2 - 1 \):
\[ \frac{d}{dx}((1+x)^2) = 2(1+x). \]

3. **Rewriting the limit**:
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{6(1+x)^5}{2(1+x)}. \]

4. **Simplifying the expression**:
We can simplify the expression:
\[ = \lim_{x \to 0} \frac{6(1+x)^5}{2(1+x)} = \lim_{x \to 0} \frac{6}{2} (1+x)^4 = \lim_{x \to 0} 3(1+x)^4. \]

5. **Evaluating the limit**:
Now substituting \( x = 0 \):
\[ 3(1+0)^4 = 3 \cdot 1 = 3. \]

Thus, the final answer is:
\[ \lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1} = 3. \]

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