Let's simplify and solve the limit:

$\lim_{x \to 0} \frac{\sin^2 x - \cos 2x + 1}{x \tan x}$

Step-by-step Solution

1. Simplify the denominator $x \tan x$: Recall that $\tan x = \frac{\sin x}{\cos x}$, so: $x \tan x = \frac{x \sin x}{\cos x}.$

2. Expand $\cos 2x$: Using the double angle identity $\cos 2x = 1 - 2\sin^2 x$, rewrite the numerator: $\sin^2 x - \cos 2x + 1 = \sin^2 x - (1 - 2\sin^2 x) + 1 = 3\sin^2 x.$

   So, the limit becomes: $\lim_{x \to 0} \frac{3\sin^2 x}{\frac{x \sin x}{\cos x}}.$

3. Simplify the fraction: Simplify the expression: $\frac{3\sin^2 x}{\frac{x \sin x}{\cos x}} = \frac{3\sin^2 x \cos x}{x \sin x}.$ Cancel one $\sin x$ from the numerator and denominator (assuming $x \neq 0$): $\frac{3\sin^2 x \cos x}{x \sin x} = \frac{3\sin x \cos x}{x}.$

4. Substitute $\sin x \approx x$ and $\cos x \approx 1$ as $x \to 0$: As $x \to 0$, $\sin x \sim x$ and $\cos x \to 1$. Substitute these approximations: $\frac{3\sin x \cos x}{x} \approx \frac{3x \cdot 1}{x} = 3.$

Final Answer:

$\boxed{3}$

Would you like further details or clarifications? 😊
Here are some related questions you might explore:

1. How does the small-angle approximation simplify trigonometric limits?
2. What happens when higher-order approximations are considered for $\sin x$ and $\cos x$?
3. How do double angle identities help in simplifying trigonometric expressions?
4. Can this method be applied to other limits involving $\tan x$?
5. How do L'Hôpital's Rule and series expansions compare for solving this limit?

Tip: Remember to always consider small-angle approximations when dealing with trigonometric functions near zero!