Question

evaluate each limit

1. \\( \lim_{x \to -3^+} \frac{x + 2}{x^2 + 5x + 6} \\)
2. \\( \lim_{x \to -2^+} \frac{3x}{x + 2} \\)
3. \\( \lim_{x \to -4} \frac{x^2}{4x + 16} \\)
4. \\( \lim_{x \to -3^-} \frac{2x}{x + 3} \\)
5. \\( \lim_{x \to -2^+} \frac{1}{x^2 - 4} \\)
6. \\( \lim_{x \to 3^-} \frac{4x}{x - 3} \\)
7. \\( \lim_{x \to 1^-} \frac{3}{x - 1} \\)
8. \\( \lim_{x \to -2^-} \frac{x + 2}{x^2 + x - 2} \\)
9. \\( \lim_{x \to -3^-} \frac{2}{x + 3} \\)
10. \\( \lim_{x \to \frac{\pi}{4}^-} 2\sec(2x) \\)
11. \\( \lim_{x \to \frac{3\pi}{4}^+} 2\tan(2x) \\)

Explanation:

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1) Step1: Factor denominator

$x^2+5x+6=(x+2)(x+3)$
Simplify: $\lim_{x
ightarrow -3^+}\frac{x+2}{(x+2)(x+3)}=\lim_{x
ightarrow -3^+}\frac{1}{x+3}$

Step2: Evaluate right-hand limit

As $x
ightarrow -3^+$, $x+3
ightarrow 0^+$, so $\frac{1}{x+3}
ightarrow +\infty$

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2) Step1: Evaluate right-hand limit

As $x
ightarrow -2^+$, $x+2
ightarrow 0^+$, numerator $3x
ightarrow -6$
$\lim_{x
ightarrow -2^+}\frac{3x}{x+2}=-\infty$

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##3) Step1: Factor denominator
$4x+16=4(x+4)$
Simplify: $\lim_{x
ightarrow -4}\frac{x^2}{4(x+4)}$

Step2: Evaluate two-sided limit

As $x
ightarrow -4^+$, $\frac{x^2}{4(x+4)}
ightarrow +\infty$; as $x
ightarrow -4^-$, $\frac{x^2}{4(x+4)}
ightarrow -\infty$. Limit does not exist.

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##4) Step1: Evaluate left-hand limit
As $x
ightarrow -3^-$, $x+3
ightarrow 0^-$, numerator $2x
ightarrow -6$
$\lim_{x
ightarrow -3^-}\frac{2x}{x+3}=+\infty$

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##5) Step1: Factor denominator
$x^2-4=(x-2)(x+2)$
Simplify: $\lim_{x
ightarrow -2^+}\frac{1}{(x-2)(x+2)}$

Step2: Evaluate right-hand limit

As $x
ightarrow -2^+$, $x+2
ightarrow 0^+$, $x-2
ightarrow -4$
$\frac{1}{(x-2)(x+2)}
ightarrow -\infty$

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##6) Step1: Evaluate left-hand limit
As $x
ightarrow 3^-$, $x-3
ightarrow 0^-$, numerator $4x
ightarrow 12$
$\lim_{x
ightarrow 3^-}\frac{4x}{x-3}=-\infty$

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##7) Step1: Evaluate left-hand limit
As $x
ightarrow 1^-$, $x-1
ightarrow 0^-$, numerator $3
ightarrow 3$
$\lim_{x
ightarrow 1^-}\frac{3}{x-1}=-\infty$

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##8) Step1: Factor denominator
$x^2+x-2=(x+2)(x-1)$
Simplify: $\lim_{x
ightarrow -2^-}\frac{x+2}{(x+2)(x-1)}=\lim_{x
ightarrow -2^-}\frac{1}{x-1}$

Step2: Evaluate left-hand limit

Substitute $x=-2$: $\frac{1}{-2-1}=-\frac{1}{3}$

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##9) Step1: Evaluate left-hand limit
As $x
ightarrow -3^-$, $x+3
ightarrow 0^-$, numerator $2
ightarrow 2$
$\lim_{x
ightarrow -3^-}\frac{2}{x+3}=-\infty$

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##10) Step1: Rewrite secant
$2\sec(2x)=\frac{2}{\cos(2x)}$

Step2: Evaluate left-hand limit

As $x
ightarrow \frac{\pi}{4}^-$, $2x
ightarrow \frac{\pi}{2}^-$, $\cos(2x)
ightarrow 0^+$
$\lim_{x
ightarrow \frac{\pi}{4}^-}2\sec(2x)=+\infty$

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##11) Step1: Evaluate right-hand limit
As $x
ightarrow \frac{3\pi}{4}^+$, $2x
ightarrow \frac{3\pi}{2}^+$, $\tan(2x)
ightarrow +\infty$
$\lim_{x
ightarrow \frac{3\pi}{4}^+}2\tan(2x)=+\infty$

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Answer:

1. $+\infty$
2. $-\infty$
3. Does not exist
4. $+\infty$
5. $-\infty$
6. $-\infty$
7. $-\infty$
8. $-\frac{1}{3}$
9. $-\infty$
10. $+\infty$
11. $+\infty$