If `lim_(xtoa)f(x)=1` and `lim_(xtoa)g(x)=oo` then `lim_(xtoa){f(x)}^(g(x))=e^(lim_(xtoa)(f(x)-1)g(x))`
`lim_(xto0)((sinx)/x)^((sinx)/(x-sinx))` is equal to

Options

A. (a) `1//e`
B. (b) `-1//e`
C. (c) `e`
D. (d) `-e`

Answer

To solve the limit problem, we will follow the steps outlined in the video transcript. The goal is to find the limit:

\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\left( \frac{\sin x}{x - \sin x} \right)} \]

### Step 1: Identify \( f(x) \) and \( g(x) \)

Let:
- \( f(x) = \frac{\sin x}{x} \)
- \( g(x) = \frac{\sin x}{x - \sin x} \)

### Step 2: Find \( \lim_{x \to 0} f(x) \)

We know from standard limits that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Thus, we have:

\[ \lim_{x \to 0} f(x) = 1 \]

### Step 3: Find \( \lim_{x \to 0} g(x) \)

Next, we need to evaluate \( g(x) \):

\[ g(x) = \frac{\sin x}{x - \sin x} \]

To find the limit as \( x \to 0 \), we can rewrite \( x - \sin x \) using the Taylor series expansion for \( \sin x \):

\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]

Thus:

\[ x - \sin x \approx x - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{6} \]

Now substituting this back into \( g(x) \):

\[ g(x) = \frac{\sin x}{\frac{x^3}{6}} \approx \frac{x}{\frac{x^3}{6}} = \frac{6}{x^2} \]

As \( x \to 0 \), \( g(x) \to \infty \).

### Step 4: Apply the limit formula

Now that we have both limits:
- \( \lim_{x \to 0} f(x) = 1 \)
- \( \lim_{x \to 0} g(x) = \infty \)

We can use the limit property:

\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x) - 1) g(x)} \]

### Step 5: Calculate \( \lim_{x \to 0} (f(x) - 1) g(x) \)

Now we need to calculate:

\[ \lim_{x \to 0} \left( \frac{\sin x}{x} - 1 \right) g(x) \]

We know:

\[ f(x) - 1 = \frac{\sin x - x}{x} = -\frac{x^3/6 + O(x^5)}{x} = -\frac{x^2}{6} + O(x^3) \]

So:

\[ (f(x) - 1) g(x) = \left(-\frac{x^2}{6}\right) \left(\frac{6}{x^2}\right) = -1 \]

Thus:

\[ \lim_{x \to 0} (f(x) - 1) g(x) = -1 \]

### Step 6: Final limit calculation

Now substituting back into the exponential limit:

\[ \lim_{x \to 0} f(x)^{g(x)} = e^{-1} = \frac{1}{e} \]

### Conclusion

Thus, the final result is:

\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\left( \frac{\sin x}{x - \sin x} \right)} = \frac{1}{e} \]