Question

\[ \lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{\sin x}+e^{\frac{1-\cos x}{x}}\right) = \underline{} \] rounded off to one decimal place.

Hint:

For limits involving \(\sin x\), \(\cos x\), and exponential functions near zero, use standard Taylor expansions.

Answer 1

Correct Answer: 1

Step 1: Use standard expansion of \(\sin x\).
\[ \sin x=x-\frac{x^3}{6}+O(x^5) \]

Step 2: Expand \(\frac{1}{\sin x}\).
\[ \frac{1}{\sin x} = \frac{1}{x}+\frac{x}{6}+O(x^3) \]

Step 3: Simplify first two terms.
\[ \frac{1}{x}-\frac{1}{\sin x} = -\frac{x}{6}+O(x^3) \]

Step 4: Take limit of first part.
\[ \lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{\sin x}\right)=0 \]

Step 5: Evaluate exponent.
Using \(1-\cos x\sim \frac{x^2}{2}\), we get
\[ \frac{1-\cos x}{x}\sim \frac{x}{2}\to 0 \]

Step 6: Evaluate exponential term.
\[ e^{\frac{1-\cos x}{x}}\to e^0=1 \]

Step 7: Final answer.
\[ 0+1=1 \] \[ \boxed{1.0} \]