Math  /  Calculus

Question6. Evaluate the following limits. a. limx4x216x4\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4} [3 marks] b. limx8x35x2+176x3+2x24x\lim _{x \rightarrow \infty} \frac{8 x^{3}-5 x^{2}+17}{6 x^{3}+2 x^{2}-4 x} [3 marks] c. limh049+h7h\lim _{h \rightarrow 0} \frac{\sqrt{49+h}-7}{h} [4 marks] d. limy2y382y27y+12\lim _{y \rightarrow-2} \frac{y^{3}-8}{2 y^{2}-7 y+12} [3 marks] e. limh042+h2h[3\lim _{h \rightarrow 0} \frac{\frac{4}{2+h}-2}{h}[3 marks ]]

Studdy Solution

STEP 1

1. We assume the limits exist and can be evaluated using algebraic manipulations or standard limit rules.
2. We will apply L'Hôpital's Rule where appropriate, which is valid when the limit yields an indeterminate form.
3. Simplification techniques like factoring and rationalizing will be used for certain limits.

STEP 2

1. Solve limit (a) by factoring and simplifying the expression.
2. Solve limit (b) by dividing the numerator and denominator by the highest power of xx in the denominator.
3. Solve limit (c) by rationalizing the numerator.
4. Solve limit (d) by direct substitution or simplifying the expression.
5. Solve limit (e) by rationalizing the numerator or using algebraic manipulation.

STEP 3

Factor the numerator in the limit limx4x216x4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}.
x216=(x4)(x+4) x^2 - 16 = (x - 4)(x + 4)

STEP 4

Substitute the factored form into the limit expression.
limx4(x4)(x+4)x4 \lim_{x \to 4} \frac{(x - 4)(x + 4)}{x - 4}

STEP 5

Cancel the common factor (x4)(x - 4).
limx4(x+4) \lim_{x \to 4} (x + 4)

STEP 6

Evaluate the limit by direct substitution.
limx4(x+4)=4+4=8 \lim_{x \to 4} (x + 4) = 4 + 4 = 8

STEP 7

For limx8x35x2+176x3+2x24x\lim_{x \to \infty} \frac{8x^3 - 5x^2 + 17}{6x^3 + 2x^2 - 4x}, divide the numerator and the denominator by x3x^3 (the highest power of xx in the denominator).
limx85x+17x36+2x4x2 \lim_{x \to \infty} \frac{8 - \frac{5}{x} + \frac{17}{x^3}}{6 + \frac{2}{x} - \frac{4}{x^2}}

STEP 8

As xx \to \infty, the terms with 1/x1/x, 1/x21/x^2, and 1/x31/x^3 tend to zero.
limx80+06+00=86=43 \lim_{x \to \infty} \frac{8 - 0 + 0}{6 + 0 - 0} = \frac{8}{6} = \frac{4}{3}

STEP 9

For limh049+h7h\lim_{h \to 0} \frac{\sqrt{49 + h} - 7}{h}, rationalize the numerator by multiplying the numerator and denominator by the conjugate 49+h+7\sqrt{49 + h} + 7.
49+h7h49+h+749+h+7=(49+h)49h(49+h+7) \frac{\sqrt{49 + h} - 7}{h} \cdot \frac{\sqrt{49 + h} + 7}{\sqrt{49 + h} + 7} = \frac{(49 + h) - 49}{h(\sqrt{49 + h} + 7)}

STEP 10

Simplify the numerator and the denominator.
hh(49+h+7)=149+h+7 \frac{h}{h(\sqrt{49 + h} + 7)} = \frac{1}{\sqrt{49 + h} + 7}

STEP 11

Evaluate the limit by direct substitution.
limh0149+h+7=149+7=17+7=114 \lim_{h \to 0} \frac{1}{\sqrt{49 + h} + 7} = \frac{1}{\sqrt{49} + 7} = \frac{1}{7 + 7} = \frac{1}{14}

STEP 12

For limy2y382y27y+12\lim_{y \to -2} \frac{y^3 - 8}{2y^2 - 7y + 12}, substitute y=2y = -2 directly to check if it results in an indeterminate form.
(2)382(2)27(2)+12=888+14+12=1634 \frac{(-2)^3 - 8}{2(-2)^2 - 7(-2) + 12} = \frac{-8 - 8}{8 + 14 + 12} = \frac{-16}{34}

STEP 13

Simplify the fraction.
1634=817 \frac{-16}{34} = -\frac{8}{17}

STEP 14

For limh042+h2h\lim_{h \to 0} \frac{\frac{4}{2 + h} - 2}{h}, simplify the numerator.
42+h2h=42(2+h)2+hh=442h2+hh=2h2+hh \frac{\frac{4}{2 + h} - 2}{h} = \frac{\frac{4 - 2(2 + h)}{2 + h}}{h} = \frac{\frac{4 - 4 - 2h}{2 + h}}{h} = \frac{\frac{-2h}{2 + h}}{h}

STEP 15

Simplify the expression further.
2h2+hh=2hh(2+h)=22+h \frac{\frac{-2h}{2 + h}}{h} = \frac{-2h}{h(2 + h)} = \frac{-2}{2 + h}

STEP 16

Evaluate the limit by direct substitution.
limh022+h=22+0=22=1 \lim_{h \to 0} \frac{-2}{2 + h} = \frac{-2}{2 + 0} = \frac{-2}{2} = -1
Solution Summary: a. limx4x216x4=8\lim_{x \to 4} \frac{x^2 - 16}{x - 4} = 8 b. limx8x35x2+176x3+2x24x=43\lim_{x \to \infty} \frac{8x^3 - 5x^2 + 17}{6x^3 + 2x^2 - 4x} = \frac{4}{3} c. limh049+h7h=114\lim_{h \to 0} \frac{\sqrt{49 + h} - 7}{h} = \frac{1}{14} d. limy2y382y27y+12=817\lim_{y \to -2} \frac{y^3 - 8}{2y^2 - 7y + 12} = -\frac{8}{17} e. limh042+h2h=1\lim_{h \to 0} \frac{\frac{4}{2 + h} - 2}{h} = -1

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