The expression in the image is:

$\lim_{{x \to 0}} \left( \frac{\sin x}{x} \cdot \frac{1}{1 - \cos x} \right)$

To evaluate this limit, let’s break it down into two parts:

1. The first part: $\frac{\sin x}{x}$
2. The second part: $\frac{1}{1 - \cos x}$

Step 1: Evaluate $\lim_{{x \to 0}} \frac{\sin x}{x}$

It is a well-known limit that: $\lim_{{x \to 0}} \frac{\sin x}{x} = 1$

Step 2: Evaluate $\lim_{{x \to 0}} \frac{1}{1 - \cos x}$

Using the trigonometric identity $1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right)$, we can rewrite this part as: $\frac{1}{1 - \cos x} = \frac{1}{2 \sin^2 \left( \frac{x}{2} \right)}$

Therefore: $\lim_{{x \to 0}} \frac{1}{1 - \cos x} = \lim_{{x \to 0}} \frac{1}{2 \sin^2 \left( \frac{x}{2} \right)}$

Substitute $u = \frac{x}{2}$, so as $x \to 0$, $u \to 0$ as well. This transforms the expression to: $\lim_{{u \to 0}} \frac{1}{2 \sin^2 u} = \frac{1}{2} \cdot \lim_{{u \to 0}} \frac{1}{\sin^2 u}$

Since $\sin u \approx u$ as $u \to 0$, this further becomes: $\frac{1}{2} \cdot \lim_{{u \to 0}} \frac{1}{u^2} = \frac{1}{2} \cdot \infty = \infty$

Conclusion

Since the second part diverges to infinity, the overall limit does not exist (it diverges to infinity).

The answer is: $\lim_{{x \to 0}} \left( \frac{\sin x}{x} \cdot \frac{1}{1 - \cos x} \right) = \infty$

Would you like more details or have any questions about the solution?

Related Questions

1. What are other common trigonometric limits to know?
2. How can we use L'Hospital's Rule to evaluate similar indeterminate forms?
3. What happens if we use series expansion to evaluate the given limit?
4. Can this limit be evaluated in polar coordinates for multivariable functions?
5. Why is $\lim_{{x \to 0}} \frac{\sin x}{x} = 1$ important in calculus?

Tip

For trigonometric limits, knowing small-angle approximations like $\sin x \approx x$ and $\cos x \approx 1 - \frac{x^2}{2}$ near $x = 0$ can simplify many calculations.