Let's go through parts (d) and (e) step-by-step.

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Part (d)

$\lim_{x \to -\infty} \frac{2x + 1}{\sqrt{x^2 - 3}}$

1. Analyze the terms: As $x \to -\infty$, both the numerator $2x + 1$ and the denominator $\sqrt{x^2 - 3}$ grow large. To evaluate this, we can factor out $x$ from both terms to simplify.

2. Rewrite the expression: Factor out $x$ from the square root in the denominator: $\frac{2x + 1}{\sqrt{x^2 - 3}} = \frac{2x + 1}{|x| \sqrt{1 - \frac{3}{x^2}}}$

   Since $x \to -\infty$, $|x| = -x$, so we have: $= \frac{2x + 1}{-x \sqrt{1 - \frac{3}{x^2}}} = \frac{2 + \frac{1}{x}}{-\sqrt{1 - \frac{3}{x^2}}}$

3. Take the limit as $x \to -\infty$:

   • As $x \to -\infty$, $\frac{1}{x} \to 0$ and $\frac{3}{x^2} \to 0$.
   • This simplifies the expression to: $= \frac{2}{-\sqrt{1}} = \frac{2}{-1} = -2$

Answer for (d): $\lim_{x \to -\infty} \frac{2x + 1}{\sqrt{x^2 - 3}} = -2$

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Part (e)

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

1. Simplify the denominator: Using the identity $1 - \cos^2 x = \sin^2 x$, we rewrite the expression as: $\lim_{x \to 0} \frac{x \sin 2x}{\sin^2 x}$

2. Simplify using trigonometric identities: We know that $\sin 2x = 2 \sin x \cos x$, so: $= \lim_{x \to 0} \frac{x \cdot 2 \sin x \cos x}{\sin^2 x}$

3. Reduce the fraction: Cancel one $\sin x$ from the numerator and denominator: $= \lim_{x \to 0} \frac{2x \cos x}{\sin x}$

4. Rewrite in terms of $\frac{\sin x}{x}$: Separate the limit as follows: $= \lim_{x \to 0} 2 \cdot \frac{x}{\sin x} \cdot \cos x$

5. Evaluate the limit:

   • $\lim_{x \to 0} \frac{x}{\sin x} = 1$
   • $\lim_{x \to 0} \cos x = 1$

   Therefore: $= 2 \cdot 1 \cdot 1 = 2$

Answer for (e): $\lim_{x \to 0} \frac{x \sin 2x}{1 - \cos^2 x} = 2$

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Let me know if you'd like more details or have further questions.

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Related Questions:

1. How do we handle limits approaching positive infinity vs. negative infinity?
2. What other trigonometric identities can simplify limit problems?
3. How does L'Hôpital's Rule apply to limit problems like these?
4. What are the key steps in rationalizing expressions to evaluate limits?
5. When should we use the $\frac{\sin x}{x} = 1$ limit property?

Tip:

When dealing with trigonometric limits, using fundamental identities and properties like $\frac{\sin x}{x} \to 1$ as $x \to 0$ can often simplify the problem significantly.