Let's go through each limit problem step-by-step.

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1. $\lim_{x \to 3} 7$

This is a constant function, so the limit will be the same as the constant value:

$\lim_{x \to 3} 7 = 7$

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2. $\lim_{x \to 1} (2x - 4x^2)$

Direct substitution:

$2(1) - 4(1)^2 = 2 - 4 = -2$

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3. $\lim_{x \to 2} \frac{2x^2 + 4x + 4}{3x - 2}$

Direct substitution first:

$\frac{2(2)^2 + 4(2) + 4}{3(2) - 2} = \frac{8 + 8 + 4}{6 - 2} = \frac{20}{4} = 5$

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4. $\lim_{x \to 4} \frac{\sqrt{x^2 + 9}}{2x}$

Evaluate by direct substitution:

$\frac{\sqrt{4^2 + 9}}{2(4)} = \frac{\sqrt{16 + 9}}{8} = \frac{\sqrt{25}}{8} = \frac{5}{8}$

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5. $\lim_{x \to 1} \frac{x^2 + 4x - 5}{x - 1}$

Direct substitution gives:

$\frac{1^2 + 4(1) - 5}{1 - 1} = \frac{0}{0}$

Since this is an indeterminate form, let's factor the numerator:

$x^2 + 4x - 5 = (x - 1)(x + 5)$

Now:

$\frac{(x - 1)(x + 5)}{x - 1} = x + 5 \quad \text{(for } x \neq 1)$

Now substitute $x = 1$:

$1 + 5 = 6$

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6. $\lim_{x \to 2} \frac{x^2 + 5x + 6}{x^2 - 3x - 10}$

Direct substitution gives:

$\frac{2^2 + 5(2) + 6}{2^2 - 3(2) - 10} = \frac{4 + 10 + 6}{4 - 6 - 10} = \frac{20}{-12} = -\frac{5}{3}$

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7. $\lim_{x \to 2} \frac{4 - x^2}{3 - \sqrt{x^2 + 5}}$

Direct substitution gives:

$\frac{4 - 2^2}{3 - \sqrt{2^2 + 5}} = \frac{4 - 4}{3 - \sqrt{9}} = \frac{0}{0}$

This is an indeterminate form. To resolve it, multiply the numerator and denominator by the conjugate:

$\frac{4 - x^2}{3 - \sqrt{x^2 + 5}} \cdot \frac{3 + \sqrt{x^2 + 5}}{3 + \sqrt{x^2 + 5}}$

Now:

$\frac{(4 - x^2)(3 + \sqrt{x^2 + 5})}{9 - (x^2 + 5)}$

Since $4 - x^2 = (2 - x)(2 + x)$, simplify the denominator:

$9 - (x^2 + 5) = 4 - x^2 = (2 - x)(2 + x)$

Now cancel out $(2 - x)(2 + x)$ from the numerator and denominator, and substitute $x = 2$:

$3 + \sqrt{2^2 + 5} = 3 + \sqrt{9} = 3 + 3 = 6$

Thus:

$\lim_{x \to 2} \frac{4 - x^2}{3 - \sqrt{x^2 + 5}} = 6$

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Final Answers

1. $7$
2. $-2$
3. $5$
4. $\frac{5}{8}$
5. $6$
6. $-\frac{5}{3}$
7. $6$

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Do you have any questions, or would you like further explanation on any of these problems?

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Related Questions

1. How do you handle indeterminate forms like $\frac{0}{0}$ in limits?
2. What are common techniques for factoring quadratic expressions?
3. How does multiplying by the conjugate help in simplifying limits?
4. Can L'Hôpital's rule be applied to all $\frac{0}{0}$ cases?
5. How can limits at infinity be handled differently from finite limits?

Tip

When direct substitution gives an indeterminate form, try simplifying the expression, factoring, or multiplying by the conjugate.