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High School Math Calculus PRECALCULUS:GRAPHICAL,...-NASTA ED.

(a) To find : The value of the limit lim y x → ∞ for the given function y = x 1 + 2 x . The value of the limit is lim y x → ∞ = 0 . Given information: The given function is y = x 1 + 2 x . Formula used: The notation lim y x → ∞ = l means that y gets arbitrarily close to l as x gets arbitrarily large. Sum rule of limits: If lim f x x → c and lim g x x → c both exist, then lim x → c f x + g x = lim x → c f x + lim x → c g x . L’Hospital’s Rule: If lim f x x → c = lim g x = 0 / ± ∞ x → c and g ′ ( x ) ≠ 0 for all x ∈ I with x ≠ c and lim f ′ x x → c lim g ′ x x → c exists, then lim x → ∞ f x g x = lim x → ∞ f ′ x g ′ x Calculation: The given function is: y = x 1 + 2 x Substitute x = ∞ in the denominator 2 ∞ means 2 × 2 × 2 × 2...... up to infinity which tends to infinity. i.e. lim 2 x = ∞ x → ∞ Calculate the limits: lim y x → ∞ = lim x → ∞ 1 + 2 x According to Sum rule of limits: lim y x → ∞ = lim 1 x → ∞ + lim 2 x x → ∞ Substitute lim 2 x = ∞ x → ∞ and lim 1 = 1 x → ∞ in the above equation to obtain the required value of the limit: lim 1 + 2 x x → ∞ = 1 + ∞ lim 1 + 2 x x → ∞ = ∞ Therefore, lim y x → ∞ = lim x → ∞ x 1 + 2 x lim y x → ∞ = ∞ ∞ Since, lim y x → ∞ = ∞ ∞ Apply L’Hospital’s rule to obtain the required limit: lim y = x → ∞ lim x → ∞ x 1 + 2 x lim y x → ∞ = lim x → ∞ d d x x lim x → ∞ d d x 1 + 2 x lim y x → ∞ = lim x → ∞ 1 lim x → ∞ 2 x ln 2 lim y x → ∞ = 1 ∞ lim y x → ∞ = 0 Therefore, the required value of the limit is 0 . (b) To find : The value of the limit lim y x → − ∞ for the given function y = x 1 + 2 x . The value of the limit is lim y x → − ∞ = − ∞ . Given information: The given function is y = x 1 + 2 x . Formula used: The notation lim y x → ∞ = l means that y gets arbitrarily close to l as x gets arbitrarily large. Sum rule of limits: If lim f x x → c and lim g x x → c both exist, then lim x → c f x + g x = lim x → c f x + lim x → c g x . Quotient rule of limits: If lim f x x → c and lim g x x → c both exist, then lim x → ∞ f x g x = lim f ( x ) x → ∞ lim g ( x ) x → ∞ . Calculation: The given function is: y = x 1 + 2 x Substitute x = − ∞ 2 − ∞ means 1 2 × 1 2 × 1 2 × 1 2 × ...... up to infinity which tends to 0 . i.e. lim 2 x = 0 x → − ∞ Calculate the limits: lim y x → − ∞ = lim x → − ∞ 1 + 2 x According to Sum rule of limits: lim 1 + 2 x x → − ∞ = lim 1 x → − ∞ + lim 2 x x → − ∞ Substitute lim 2 x = 0 x → − ∞ and lim 1 = 1 x → − ∞ in the above equation to obtain the required value of the limit: lim 1 + 2 x x → − ∞ = 1 + 0 lim 1 + 2 x x → − ∞ = 1 Calculate the required limit: lim y = x → − ∞ lim x → − ∞ x 1 + 2 x Apply Quotient rule of limits: lim y = x → − ∞ lim x x → − ∞ lim ( 1 + 2 x ) x → − ∞ Substitute lim 1 + 2 x x → − ∞ = 1 and lim x x → − ∞ = − ∞ in the above equation to obtain the required value: lim y = x → − ∞ − ∞ 1 lim y = x → − ∞ − ∞ Therefore, the required value of the limit is − ∞ .

(a) To find : The value of the limit lim y x → ∞ for the given function y = x 1 + 2 x . The value of the limit is lim y x → ∞ = 0 . Given information: The given function is y = x 1 + 2 x . Formula used: The notation lim y x → ∞ = l means that y gets arbitrarily close to l as x gets arbitrarily large. Sum rule of limits: If lim f x x → c and lim g x x → c both exist, then lim x → c f x + g x = lim x → c f x + lim x → c g x . L’Hospital’s Rule: If lim f x x → c = lim g x = 0 / ± ∞ x → c and g ′ ( x ) ≠ 0 for all x ∈ I with x ≠ c and lim f ′ x x → c lim g ′ x x → c exists, then lim x → ∞ f x g x = lim x → ∞ f ′ x g ′ x Calculation: The given function is: y = x 1 + 2 x Substitute x = ∞ in the denominator 2 ∞ means 2 × 2 × 2 × 2...... up to infinity which tends to infinity. i.e. lim 2 x = ∞ x → ∞ Calculate the limits: lim y x → ∞ = lim x → ∞ 1 + 2 x According to Sum rule of limits: lim y x → ∞ = lim 1 x → ∞ + lim 2 x x → ∞ Substitute lim 2 x = ∞ x → ∞ and lim 1 = 1 x → ∞ in the above equation to obtain the required value of the limit: lim 1 + 2 x x → ∞ = 1 + ∞ lim 1 + 2 x x → ∞ = ∞ Therefore, lim y x → ∞ = lim x → ∞ x 1 + 2 x lim y x → ∞ = ∞ ∞ Since, lim y x → ∞ = ∞ ∞ Apply L’Hospital’s rule to obtain the required limit: lim y = x → ∞ lim x → ∞ x 1 + 2 x lim y x → ∞ = lim x → ∞ d d x x lim x → ∞ d d x 1 + 2 x lim y x → ∞ = lim x → ∞ 1 lim x → ∞ 2 x ln 2 lim y x → ∞ = 1 ∞ lim y x → ∞ = 0 Therefore, the required value of the limit is 0 . (b) To find : The value of the limit lim y x → − ∞ for the given function y = x 1 + 2 x . The value of the limit is lim y x → − ∞ = − ∞ . Given information: The given function is y = x 1 + 2 x . Formula used: The notation lim y x → ∞ = l means that y gets arbitrarily close to l as x gets arbitrarily large. Sum rule of limits: If lim f x x → c and lim g x x → c both exist, then lim x → c f x + g x = lim x → c f x + lim x → c g x . Quotient rule of limits: If lim f x x → c and lim g x x → c both exist, then lim x → ∞ f x g x = lim f ( x ) x → ∞ lim g ( x ) x → ∞ . Calculation: The given function is: y = x 1 + 2 x Substitute x = − ∞ 2 − ∞ means 1 2 × 1 2 × 1 2 × 1 2 × ...... up to infinity which tends to 0 . i.e. lim 2 x = 0 x → − ∞ Calculate the limits: lim y x → − ∞ = lim x → − ∞ 1 + 2 x According to Sum rule of limits: lim 1 + 2 x x → − ∞ = lim 1 x → − ∞ + lim 2 x x → − ∞ Substitute lim 2 x = 0 x → − ∞ and lim 1 = 1 x → − ∞ in the above equation to obtain the required value of the limit: lim 1 + 2 x x → − ∞ = 1 + 0 lim 1 + 2 x x → − ∞ = 1 Calculate the required limit: lim y = x → − ∞ lim x → − ∞ x 1 + 2 x Apply Quotient rule of limits: lim y = x → − ∞ lim x x → − ∞ lim ( 1 + 2 x ) x → − ∞ Substitute lim 1 + 2 x x → − ∞ = 1 and lim x x → − ∞ = − ∞ in the above equation to obtain the required value: lim y = x → − ∞ − ∞ 1 lim y = x → − ∞ − ∞ Therefore, the required value of the limit is − ∞ .

PRECALCULUS:GRAPHICAL,...-NASTA ED.

PRECALCULUS:GRAPHICAL,...-NASTA ED.

10th Edition

ISBN: 9780134672090

Author: Demana

Publisher: PEARSON

See similar textbooks

Expert Solution & Answer

Chapter 11.3, Problem 50E

Solution

(a)

To find: The value of the limit $limyx→∞$ for the given function $y=x1+2x$ .

The value of the limit is $limyx→∞=0$ .

Given information:

The given function is $y=x1+2x$ .

Formula used:

The notation $limyx→∞=l$ means that $y$ gets arbitrarily close to $l$ as $x$ gets arbitrarily large.

Sum rule of limits:

If $limfxx→c$ and $limgxx→c$ both exist, then $limx→cfx+gx=limx→cfx+limx→cgx$ .

L’Hospital’s Rule:

If $limfxx→c=limgx=0/±∞x→c$ and $g′(x)≠0$ for all $x∈I$ with $x≠c$ and $limf′xx→climg′xx→c$ exists, then

$limx→∞fxgx=limx→∞f′xg′x$

Calculation:

The given function is: $y=x1+2x$

Substitute $x=∞$ in the denominator

$2∞$ means $2×2×2×2......$ up to infinity which tends to infinity.

i.e. $lim2x=∞x→∞$

Calculate the limits:

$limyx→∞=limx→∞1+2x$

According to Sum rule of limits:

$limyx→∞=lim1x→∞+lim2xx→∞$

Substitute $lim2x=∞x→∞$ and $lim1=1x→∞$

in the above equation to obtain the required value of the limit:

$lim1+2xx→∞=1+∞lim1+2xx→∞=∞$

Therefore,

$limyx→∞=limx→∞x1+2xlimyx→∞=∞∞$

Since, $limyx→∞=∞∞$

Apply L’Hospital’s rule to obtain the required limit:

$limy=x→∞limx→∞x1+2xlimyx→∞=limx→∞ddxxlimx→∞ddx1+2xlimyx→∞=limx→∞1limx→∞2xln2limyx→∞=1∞limyx→∞=0$

Therefore, the required value of the limit is $0$ .

(b)

To find: The value of the limit $limyx→−∞$ for the given function $y=x1+2x$ .

The value of the limit is $limyx→−∞=−∞$ .

Given information:

The given function is $y=x1+2x$ .

Formula used:

The notation $limyx→∞=l$ means that $y$ gets arbitrarily close to $l$ as $x$ gets arbitrarily large.

Sum rule of limits:

If $limfxx→c$ and $limgxx→c$ both exist, then $limx→cfx+gx=limx→cfx+limx→cgx$ .

Quotient rule of limits:

If $limfxx→c$ and $limgxx→c$ both exist, then $limx→∞fxgx=limf(x)x→∞limg(x)x→∞$ .

Calculation:

The given function is: $y=x1+2x$

Substitute $x=−∞$

$2−∞$ means $12×12×12×12×......$ up to infinity which tends to $0$ .

i.e. $lim2x=0x→−∞$

Calculate the limits:

$limyx→−∞=limx→−∞1+2x$

According to Sum rule of limits:

$lim1+2xx→−∞=lim1x→−∞+lim2xx→−∞$

Substitute $lim2x=0x→−∞$ and $lim1=1x→−∞$

in the above equation to obtain the required value of the limit:

$lim1+2xx→−∞=1+0lim1+2xx→−∞=1$

Calculate the required limit:

$limy=x→−∞limx→−∞x1+2x$

Apply Quotient rule of limits:

$limy=x→−∞limxx→−∞lim(1+2x)x→−∞$

Substitute $lim1+2xx→−∞=1$ and $limxx→−∞=−∞$

in the above equation to obtain the required value:

$limy=x→−∞−∞1limy=x→−∞−∞$

Therefore, the required value of the limit is $−∞$ .

Chapter 11.3, Problem 49E

Chapter 11.3, Problem 51E

Chapter 11 Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

Ch. 11.1 - Prob. 1QR Ch. 11.1 - Prob. 2QR Ch. 11.1 - Prob. 3QR Ch. 11.1 - Prob. 4QR Ch. 11.1 - Prob. 5QR Ch. 11.1 - Prob. 6QR Ch. 11.1 - Prob. 7QR Ch. 11.1 - Prob. 8QR Ch. 11.1 - Prob. 9QR Ch. 11.1 - Prob. 10QR

Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E Ch. 11.1 - Prob. 11E Ch. 11.1 - Prob. 12E Ch. 11.1 - Prob. 13E Ch. 11.1 - Prob. 14E Ch. 11.1 - Prob. 15E Ch. 11.1 - Prob. 16E Ch. 11.1 - Prob. 17E Ch. 11.1 - Prob. 18E Ch. 11.1 - Prob. 19E Ch. 11.1 - Prob. 20E Ch. 11.1 - Prob. 21E Ch. 11.1 - Prob. 22E Ch. 11.1 - Prob. 23E Ch. 11.1 - Prob. 24E Ch. 11.1 - Prob. 25E Ch. 11.1 - Prob. 26E Ch. 11.1 - Prob. 27E Ch. 11.1 - Prob. 28E Ch. 11.1 - Prob. 29E Ch. 11.1 - Prob. 30E Ch. 11.1 - Prob. 31E Ch. 11.1 - Prob. 32E Ch. 11.1 - Prob. 33E Ch. 11.1 - Prob. 34E Ch. 11.1 - Prob. 35E Ch. 11.1 - Prob. 36E Ch. 11.1 - Prob. 37E Ch. 11.1 - Prob. 38E Ch. 11.1 - Prob. 39E Ch. 11.1 - Prob. 40E Ch. 11.1 - Prob. 41E Ch. 11.1 - Prob. 42E Ch. 11.1 - Prob. 43E Ch. 11.1 - Prob. 44E Ch. 11.1 - Prob. 45E Ch. 11.1 - Prob. 46E Ch. 11.1 - Prob. 47E Ch. 11.1 - Prob. 48E Ch. 11.1 - 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Prob. 5E Ch. 11.3 - Prob. 6E Ch. 11.3 - Prob. 7E Ch. 11.3 - Prob. 8E Ch. 11.3 - Prob. 9E Ch. 11.3 - Prob. 10E Ch. 11.3 - Prob. 11E Ch. 11.3 - Prob. 12E Ch. 11.3 - Prob. 13E Ch. 11.3 - Prob. 14E Ch. 11.3 - Prob. 15E Ch. 11.3 - Prob. 16E Ch. 11.3 - Prob. 17E Ch. 11.3 - Prob. 18E Ch. 11.3 - Prob. 19E Ch. 11.3 - Prob. 20E Ch. 11.3 - Prob. 21E Ch. 11.3 - Prob. 22E Ch. 11.3 - Prob. 23E Ch. 11.3 - Prob. 24E Ch. 11.3 - Prob. 25E Ch. 11.3 - Prob. 26E Ch. 11.3 - Prob. 27E Ch. 11.3 - Prob. 28E Ch. 11.3 - Prob. 29E Ch. 11.3 - Prob. 30E Ch. 11.3 - Prob. 31E Ch. 11.3 - Prob. 32E Ch. 11.3 - Prob. 33E Ch. 11.3 - Prob. 34E Ch. 11.3 - Prob. 35E Ch. 11.3 - Prob. 36E Ch. 11.3 - Prob. 37E Ch. 11.3 - Prob. 38E Ch. 11.3 - Prob. 39E Ch. 11.3 - Prob. 40E Ch. 11.3 - Prob. 41E Ch. 11.3 - Prob. 42E Ch. 11.3 - Prob. 43E Ch. 11.3 - Prob. 44E Ch. 11.3 - Prob. 45E Ch. 11.3 - Prob. 46E Ch. 11.3 - Prob. 47E Ch. 11.3 - Prob. 48E Ch. 11.3 - Prob. 49E Ch. 11.3 - Prob. 50E Ch. 11.3 - Prob. 51E Ch. 11.3 - Prob. 52E Ch. 11.3 - 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Prob. 9QR Ch. 11.4 - Prob. 10QR Ch. 11.4 - Prob. 1E Ch. 11.4 - Prob. 2E Ch. 11.4 - Prob. 3E Ch. 11.4 - Prob. 4E Ch. 11.4 - Prob. 5E Ch. 11.4 - Prob. 6E Ch. 11.4 - Prob. 7E Ch. 11.4 - Prob. 8E Ch. 11.4 - Prob. 9E Ch. 11.4 - Prob. 10E Ch. 11.4 - Prob. 11E Ch. 11.4 - Prob. 12E Ch. 11.4 - Prob. 13E Ch. 11.4 - Prob. 14E Ch. 11.4 - Prob. 15E Ch. 11.4 - Prob. 16E Ch. 11.4 - Prob. 17E Ch. 11.4 - Prob. 18E Ch. 11.4 - Prob. 19E Ch. 11.4 - Prob. 20E Ch. 11.4 - Prob. 21E Ch. 11.4 - Prob. 22E Ch. 11.4 - Prob. 23E Ch. 11.4 - Prob. 24E Ch. 11.4 - Prob. 25E Ch. 11.4 - Prob. 26E Ch. 11.4 - Prob. 27E Ch. 11.4 - Prob. 28E Ch. 11.4 - Prob. 29E Ch. 11.4 - Prob. 30E Ch. 11.4 - Prob. 31E Ch. 11.4 - Prob. 32E Ch. 11.4 - Prob. 33E Ch. 11.4 - Prob. 34E Ch. 11.4 - Prob. 35E Ch. 11.4 - Prob. 36E Ch. 11.4 - Prob. 37E Ch. 11.4 - Prob. 38E Ch. 11.4 - Prob. 39E Ch. 11.4 - Prob. 40E Ch. 11.4 - Prob. 41E Ch. 11.4 - Prob. 42E Ch. 11.4 - Prob. 43E Ch. 11.4 - Prob. 44E Ch. 11.4 - Prob. 45E Ch. 11.4 - Prob. 46E Ch. 11.4 - 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