If $f:R\rightarrow R$ such that $f(1)=2$ and $f^{\prime}(1)=6$, then $lim_{x\rightarrow0}[\frac{f(1+x)}{f(1)}]^{\frac{1}{x}}=$

Q: If $f:R\rightarrow R$ such that $f(1)=2$ and $f^{\prime}(1)=6$, then $lim_{x\rightarrow0}[\frac{f(1+x)}{f(1)}]^{\frac{1}{x}}=$

To solve the problem, we need to evaluate the limit: \[ \lim_{x \rightarrow 0} \left[\frac{f(1+x)}{f(1)}\right]^{\frac{1}{x}} \] Given that $ f(1) = 2 $ and $ f'(1) = 6 $, we can use the concept of the derivative to approximate $ f(1+x) $ using the linear approximation (or the first-order Taylor expansion) around $ x = 0 $: \[ f(1+x) \approx f(1) + f'(1) \cdot x = 2 + 6x \] Substituting this approximation into the limit expression, we have: \[ \lim_{x \rightarrow 0} \left[\frac{2 + 6x}{2}\right]^{\frac{1}{x}} \] Simplifying inside the bracket: \[ \frac{2 + 6x}{2} = 1 + 3x \] Thus, the limit becomes: \[ \lim_{x \rightarrow 0} (1 + 3x)^{\frac{1}{x}} \] This expression is a standard limit that evaluates to the exponential function: \[ \lim_{x \rightarrow 0} (1 + 3x)^{\frac{1}{x}} = e^3 \] Therefore, the correct answer is $ e^3 $, which corresponds to Option 4. ### Explanation for Other Options: 1. **Option 1 (1):** This would be the result if the function inside the limit did not change with $ x $, i.e., if $ f(1+x) = f(1) $ for all $ x $. However, since $ f'(1) = 6 $, the function does change, and the limit is not 1. 2. **Option 2 (e):** This would be the result if the linear term in the approximation was $ x $ instead of $ 3x $. Since the derivative $ f'(1) = 6 $ leads to a linear term of $ 3x $, the limit evaluates to $ e^3 $, not $ e $. 3. **Option 3 ($ e^2 $):** This would be the result if the linear term in the approximation was $ 2x $. However, the correct linear term is $ 3x $, leading to the limit being $ e^3 $. Thus, the correct answer is indeed Option 4, $ e^3 $, based on the given conditions and the standard limit evaluation.

Question

Accepted Answer

To solve the problem, we need to evaluate the limit:

$$
\lim_{x \rightarrow 0} \left[\frac{f(1+x)}{f(1)}\right]^{\frac{1}{x}}
$$

Given that $ f(1) = 2 $ and $ f'(1) = 6 $, we can use the concept of the derivative to approximate $ f(1+x) $ using the linear approximation (or the first-order Taylor expansion) around $ x = 0 $:

$$
f(1+x) \approx f(1) + f'(1) \cdot x = 2 + 6x
$$

Substituting this approximation into the limit expression, we have:

$$
\lim_{x \rightarrow 0} \left[\frac{2 + 6x}{2}\right]^{\frac{1}{x}}
$$

Simplifying inside the bracket:

$$
\frac{2 + 6x}{2} = 1 + 3x
$$

Thus, the limit becomes:

$$
\lim_{x \rightarrow 0} (1 + 3x)^{\frac{1}{x}}
$$

This expression is a standard limit that evaluates to the exponential function:

$$
\lim_{x \rightarrow 0} (1 + 3x)^{\frac{1}{x}} = e^3
$$

Therefore, the correct answer is $ e^3 $, which corresponds to Option 4.

### Explanation for Other Options:

1. **Option 1 (1):** This would be the result if the function inside the limit did not change with $ x $, i.e., if $ f(1+x) = f(1) $ for all $ x $. However, since $ f'(1) = 6 $, the function does change, and the limit is not 1.

2. **Option 2 (e):** This would be the result if the linear term in the approximation was $ x $ instead of $ 3x $. Since the derivative $ f'(1) = 6 $ leads to a linear term of $ 3x $, the limit evaluates to $ e^3 $, not $ e $.

3. **Option 3 ($ e^2 $):** This would be the result if the linear term in the approximation was $ 2x $. However, the correct linear term is $ 3x $, leading to the limit being $ e^3 $.

Thus, the correct answer is indeed Option 4, $ e^3 $, based on the given conditions and the standard limit evaluation.