Question

\( \lim_{x \to 0} \frac{e^{\sin x} - 1}{x} \) is equal to

• A. $0$
• B. $e$
• C. $1$
• D. Does not exist

Hint:

For $x \to 0$, use $\sin x \approx x$ and $e^x \approx 1 + x$.

Answer 1

The Correct Option is C

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

\(\lim_{x \to 0} \frac{e^{\sin x} - 1}{x}\)

We know from the limit properties that to handle expressions of the form \(\lim_{x \to 0} \frac{f(x) - f(0)}{x}\), we can utilize L'Hôpital's Rule, which is applicable for limits resulting in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

1. First, let's verify the indeterminate form:
   • As \(x \to 0\), \(\sin x \to 0\).
   • \(e^{\sin x} - 1\) as \(x \to 0\) becomes \(e^0 - 1 = 1 - 1 = 0\).
   • Thus, the limit transforms into \(\frac{0}{0}\), an indeterminate form.
2. Apply L'Hôpital's Rule, which tells us to differentiate the numerator and the denominator separately:
   • The derivative of the numerator \(e^{\sin x} - 1\) is \(e^{\sin x} \cdot \cos x\) (using chain rule).
   • The derivative of the denominator \(x\) is \(1\).
3. Substitute these derivatives back into the limit:
   • \(\lim_{x \to 0} \frac{e^{\sin x} \cdot \cos x}{1} = \lim_{x \to 0} e^{\sin x} \cdot \cos x\)
   • Since \(\sin x \to 0\) and \(\cos x \to 1\) as \(x \to 0\), we have: \( e^{0} \cdot 1 = 1 \)

Therefore, \(\lim_{x \to 0} \frac{e^{\sin x} - 1}{x} = 1\)

Thus, the correct answer is \(1\).