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: Split the problem.
We look at $\Big(\frac1x-\frac{1}{\sin x}\Big)$ and the term $e^{(1-\cos x)/x}$ separately.

Step 2: Expand $\sin x$.
Using $\sin x=x-\frac{x^3}{6}+\cdots$, we get $\frac{1}{\sin x}=\frac1x+\frac{x}{6}+\cdots$.

Step 3: First bracket.
So $\frac1x-\frac{1}{\sin x}=-\frac{x}{6}+\cdots$, which goes to $0$ as $x\to 0$.

Step 4: Look at the exponent.
Since $1-\cos x\approx \frac{x^2}{2}$, the exponent $\frac{1-\cos x}{x}\approx \frac{x}{2}\to 0$.

Step 5: The exponential term.
So $e^{(1-\cos x)/x}\to e^{0}=1$.

Step 6: Add the pieces.
The total limit is $0+1=1$.
\[ \boxed{1.0} \]