Question

abstract. practice problems for lecture
problem. 1 : determine if the limit approaches a finite number, $\infty$, $-\infty$, or does not exist. (if the limit does not exist, write dne)
$\lim \limits_{x \to 10} \frac{3 x^2 - 42 x + 120}{x - 10} = $
problem. 2 : determine if the limit approaches a finite number, $\infty$, $-\infty$, or does not exist. (if the limit does not exist, write dne)
$\lim \limits_{x \to 4} \frac{2 x - 8}{x^2 - 10 x + 24} = $
problem. 3 : determine if the limit approaches a finite number, $\infty$, $-\infty$, or does not exist. (if the limit does not exist, write dne)
$\lim \limits_{x \to -8} \frac{3 x^2 + 9 x - 120}{x - 5} = $
problem. 4 : determine if the limit approaches a finite number, $\infty$, $-\infty$, or does not exist. (if the limit does not exist, write dne)
$\lim \limits_{x \to -9} \frac{2 x + 20}{x^2 + 19 x + 90} = $

Explanation:

Step1: Factor numerator (Problem 1)

$3x^2 - 42x + 120 = 3(x-10)(x-4)$

Step2: Simplify the expression (Problem 1)

$\lim_{x \to 10} \frac{3(x-10)(x-4)}{x-10} = \lim_{x \to 10} 3(x-4)$

Step3: Evaluate the limit (Problem 1)

$3(10-4) = 18$

Step1: Factor numerator/denominator (Problem 2)

$2x-8=2(x-4)$; $x^2-10x+24=(x-4)(x-6)$

Step2: Simplify the expression (Problem 2)

$\lim_{x \to 4} \frac{2(x-4)}{(x-4)(x-6)} = \lim_{x \to 4} \frac{2}{x-6}$

Step3: Evaluate the limit (Problem 2)

$\frac{2}{4-6} = -1$

Step1: Substitute $x=-8$ directly (Problem 3)

Denominator: $-8-5=-13
eq 0$, so no simplification needed

Step2: Compute numerator (Problem 3)

$3(-8)^2 +9(-8)-120=3(64)-72-120=192-192= -33$

Step3: Evaluate the limit (Problem 3)

$\frac{-33}{-13} = \frac{33}{13}$

Step1: Factor numerator/denominator (Problem 4)

$2x+20=2(x+10)$; $x^2+19x+90=(x+9)(x+10)$

Step2: Simplify the expression (Problem 4)

$\lim_{x \to -9} \frac{2(x+10)}{(x+9)(x+10)} = \lim_{x \to -9} \frac{2}{x+9}$
Wait, correction: Re-evaluate substitution:
Wait, $x \to -9$, denominator $(-9)+9=0$, numerator $2(-9+10)=2(1)=2$
Wait no, re-factor denominator correctly: $x^2+19x+90=(x+9)(x+10)$, yes. When $x \to -9$, the simplified limit is $\lim_{x \to -9} \frac{2}{x+9}$. As $x \to -9^+$, $x+9 \to 0^+$, limit is $+\infty$; as $x \to -9^-$, $x+9 \to 0^-$, limit is $-\infty$. Wait no, original problem: $\lim_{x \to -9} \frac{2x+20}{x^2+19x+90}$
Wait re-calculate numerator at $x=-9$: $2(-9)+20=-18+20=2$. Denominator: $(-9)^2+19(-9)+90=81-171+90=0$. Now check left and right limits:
As $x \to -9^+$, $x+9 \to 0^+$, so $\frac{2}{x+9} \to +\infty$
As $x \to -9^-$, $x+9 \to 0^-$, so $\frac{2}{x+9} \to -\infty$
Since left and right limits are not equal, the limit DNE? Wait no, wait I made a mistake in factoring:
Wait $2x+20=2(x+10)$, $x^2+19x+90=(x+9)(x+10)$. So when $x
eq -10$, we can cancel $x+10$. But $x \to -9$, so $x
eq -10$, so simplification is valid. Then $\lim_{x \to -9} \frac{2}{x+9}$. Since left limit is $-\infty$ and right limit is $+\infty$, the two-sided limit does not exist.

Wait correction for Problem 4:

Step1: Factor numerator/denominator (Problem 4)

$2x+20=2(x+10)$; $x^2+19x+90=(x+9)(x+10)$

Step2: Simplify the expression (Problem 4)

$\lim_{x \to -9} \frac{2(x+10)}{(x+9)(x+10)} = \lim_{x \to -9} \frac{2}{x+9}$ (for $x
eq -10$)

Step3: Evaluate left/right limits (Problem 4)

Left limit ($x \to -9^-$): $\frac{2}{0^-} = -\infty$; Right limit ($x \to -9^+$): $\frac{2}{0^+} = +\infty$

Step4: Conclude limit status (Problem 4)

Left $
eq$ right limit, so limit DNE

Final corrected answers:
1. $18$
2. $-1$
3. $\frac{33}{13}$
4. DNE

Revised Answer:

1. $18$
2. $-1$
3. $\frac{33}{13}$
4. DNE

Revised Explanation:

Problem 1

Step1: Factor numerator

$3x^2 - 42x + 120 = 3(x-10)(x-4)$

Step2: Cancel common factor

$\lim_{x \to 10} \frac{3(x-10)(x-4)}{x-10} = \lim_{x \to 10} 3(x-4)$

Step3: Compute limit

$3(10-4) = 18$

Problem 2

Step1: Factor numerator/denominator

$2x-8=2(x-4)$; $x^2-10x+24=(x-4)(x-6)$

Step2: Cancel common factor

$\lim_{x \to 4} \frac{2(x-4)}{(x-4)(x-6)} = \lim_{x \to 4} \frac{2}{x-6}$

Step3: Compute limit

$\frac{2}{4-6} = -1$

Problem 3

Step1: Substitute $x=-8$ directly

Denominator: $-8-5=-13
eq 0$

Step2: Calculate numerator

$3(-8)^2 +9(-8)-120=192-72-120=-33$

Step3: Compute limit

$\frac{-33}{-13} = \frac{33}{13}$

Problem 4

Step1: Factor numerator/denominator

$2x+20=2(x+10)$; $x^2+19x+90=(x+9)(x+10)$

Step2: Cancel common factor

$\lim_{x \to -9} \frac{2(x+10)}{(x+9)(x+10)} = \lim_{x \to -9} \frac{2}{x+9}$

Step3: Evaluate one-sided limits

Left limit: $\lim_{x \to -9^-} \frac{2}{x+9} = -\infty$; Right limit: $\lim_{x \to -9^+} \frac{2}{x+9} = +\infty$

Step4: Determine limit existence

One-sided limits differ, so limit DNE

Answer:

1. $18$
2. $-1$
3. $\frac{33}{13}$
4. $-2$