Exercise 1

f(x) = \frac { { x }^{ 2 } + 2 }{ x - 2 }

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Exercise 2

f(x) = \frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } }

Exercise 3

f(x) = \frac { { x }^{ 4 } + 1 }{ { x }^{ 2 } }

Exercise 4

f(x) = \frac { { x }^{ 2 } }{ 2 - x }

Exercise 5

f(x) = \frac { x }{ 1 + { x }^{ 2 } }

Exercise 6

f(x) = \frac { { x }^{ 2 } - 3x + 2 }{ { x }^{ 2 } + 1 }

Exercise 7

f(x) = \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } }

Exercise 8

f(x) = { e }^{ \frac { 1 }{ x } }

Exercise 9

f(x) = (x - 1){ e }^{ -x }

Exercise 10

f(x) = \frac { 1 }{ 2 \sqrt { 2 \pi }} { e }^{ - \frac { 1 }{ 2 } { x }^{ 2 } }

Exercise 11

f(x) = \frac { \ln { x } }{ x }

 

Solution of exercise 1

f(x) = \frac { { x }^{ 2 } + 2 }{ x - 2 }

Horizontal asymptotes

\lim_{ x \rightarrow \infty } \frac { { x }^{ 2 } + 2 }{ x - 2 } = \infty

Vertical asymptotes.

\lim_{ x \rightarrow 2 } \frac { { x }^{ 2 } + 2 }{ x - 2 } = \infty

Oblique asymptotes.

m = \lim_{ x \rightarrow  \infty } \frac { \frac { { x }^{ 2 } + 2 }{ x - 2 } }{ x } = \lim_{ x \rightarrow \infty } \frac { { x }^{ 2 } + 2 }{ { x }^{ 2 } - 2x } = 1

n = \lim_{ x \rightarrow \infty } ( \frac { { x }^{ 2 } + 2 }{ x - 2 } - (1 \times x) ) = \lim_{ x \rightarrow \infty } \frac { 2x + 2 }{ x - 2 } = 2

y = x + 2

 

Solution of exercise 2

f(x) = \frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } }

Horizontal asymptotes

\lim_{ x \rightarrow \infty } \frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } } = \infty \qquad No

Vertical asymptotes.

\lim_{ x \rightarrow 1 } \frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } } = \infty \qquad x = 1

Oblique asymptotes.

m = \lim_{ x \rightarrow \infty } \frac { \frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } } }{ x } = 1

n = \lim_{ x \rightarrow \infty } [\frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } } - x] = 2

y = x + 2

 

Solution of exercise 3

f(x) = \frac { { x }^{ 4 } + 1 }{ { x }^{ 2 } }

Horizontal asymptotes

\lim_{ x \rightarrow \infty } f(x) = \frac { { x }^{ 4 } + 1 }{ { x }^{ 2 } } = \infty \qquad No

Vertical asymptotes.

\lim_{ x \rightarrow 0 } f(x) = \frac { { x }^{ 4 } + 1 }{ { x }^{ 2 } } = \infty \qquad x = 0

Oblique asymptotes.

m = \lim_{ x \rightarrow \infty } \frac { \frac { { x }^{ 4 } + 1 }{ { x }^{ 2 } } }{ x } = \lim_{ x \rightarrow \infty } \frac { { x }^{ 4 } + 1 }{ { x }^{ 3 } } = \infty \qquad No

 

Solution of exercise 4

f(x) = \frac { { x }^{ 2 } }{ 2 - x }

Horizontal asymptotes

\lim_{ x \rightarrow \infty } \frac { { x }^{ 2 } }{ 2 - x } = \infty \qquad No

Vertical asymptotes.

\lim_{ x \rightarrow 2 } \frac { { x }^{ 2 } }{ 2 - x } = \infty \qquad x = 2

Oblique asymptotes.

m = \lim_{ x \rightarrow \infty } \frac { \frac { { x }^{ 2 } }{ 2 - x } }{ x } = -1

n = \lim_{ x \rightarrow \infty } (\frac { { x }^{ 2 } }{ 2 - x } + x) = -2

y = -x -2

 

Solution of exercise 5

f(x) = \frac { x }{ 1 + { x }^{ 2 } }

Horizontal asymptotes

\lim_{ x \rightarrow \infty } \frac { x }{ 1 + { x }^{ 2 } } = 0 \qquad y = 0

No vertical or oblique asymptote.

 

Solution of exercise 6

f(x) = \frac { { x }^{ 2 } - 3x + 2 }{ { x }^{ 2 } + 1 }

Horizontal asymptotes

\lim_{ x \rightarrow \infty } \frac { { x }^{ 2 } - 3x + 2 }{ { x }^{ 2 } + 1 } = 1 \qquad y = 1

No vertical or oblique asymptote.

 

Solution of exercise 7

f(x) = \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } }

{ x }^{ 2 } - 1 } > 0 \qquad x = \pm 1 \qquad D = (- \infty, -1)U(1, \infty)

Horizontal asymptotes.

\lim_{x \rightarrow \infty } \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } } = \infty

No Horizontal Asymptote.

Vertical asymptotes.

\lim_{x \rightarrow -1 } \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } } = \infty \qquad x = -1

\lim_{x \rightarrow 1 } \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } } = \infty \qquad x = 1

Oblique asymptotes.

m = \lim_{x \rightarrow \infty } \frac { \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } } }{ x } = \lim_{x \rightarrow \infty } \frac { { x }^{ 2 } }{ x \sqrt { { x }^{ 2 } - 1 } } = \lim_{x \rightarrow \infty } \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 4 } - { x }^{ 2 } } } = 1

n = \lim_{x \rightarrow \infty } (\frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } } - x) = \lim_{x \rightarrow \infty } \frac { { x }^{ 2 } - x \sqrt { { x }^{ 2 } - 1 } }{ \sqrt { { x }^{ 2 } - 1 } }

= \lim_{x \rightarrow \infty } \frac { { x }^{ 2 } - \sqrt { { x }^{ 4 } - { x }^{ 2 } } }{ \sqrt { { x }^{ 2 } - 1 } } = \lim_{x \rightarrow \infty } \frac { ({ x }^{ 2 } - \sqrt { { x }^{ 4 } - { x }^{ 2 } })( { x }^{ 2 } + \sqrt { { x }^{ 4 } - { x }^{ 2 } } ) }{ \sqrt { { x }^{ 2 } - 1 } ({ x }^{ 2 } + \sqrt { { x }^{ 4 } - { x }^{ 2 } }) }

\lim_{x \rightarrow \infty } \frac { { x }^{ 4 } - ({ x }^{ 4 } - { x }^{ 2 }) }{ \sqrt { { x }^{ 2 } - 1 } ({ x }^{ 2 } + \sqrt { { x }^{ 4 } - { x }^{ 2 } }) } = \lim_{x \rightarrow \infty } \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 2 } - 1 } ({ x }^{ 2 } + \sqrt { { x }^{ 4 } - { x }^{ 2 } }) } = 0

y = x

 

Solution of exercise 8

f(x) = { e }^{ \frac { 1 }{ x } }

Horizontal asymptotes

\lim_{x \rightarrow \infty } { e }^{ \frac { 1 }{ x } } = { e }^{ 0 } = 1 \qquad y = 1

Vertical asymptotes.

\lim_{x \rightarrow 0 } { e }^{ \frac { 1 }{ x } } =\infty \qquad x = 0

 

Solution of exercise 9

f(x) = (x - 1){ e }^{ -x }

Horizontal asymptotes

\lim_{x \rightarrow \infty } (x - 1){ e }^{ -x } = \lim_{x \rightarrow \infty } \frac { x - 1 }{ { e }^{ -x } } = 0 \qquad y = 0

No vertical or oblique asymptote.

 

Solution of exercise 10

f(x) = \frac { 1 }{ 2 \sqrt { 2 \pi }} { e }^{ - \frac { 1 }{ 2 } { x }^{ 2 } }

Horizontal asymptotes

\lim_{x \rightarrow \infty } \frac { 1 }{ 2 \sqrt { 2 \pi }} { e }^{ - \frac { 1 }{ 2 } { x }^{ 2 } } =  0 \qquad y = 0

No vertical or oblique asymptote.

 

Solution of exercise 11

f(x) = \frac { \ln { x } }{ x }

Horizontal asymptotes

\lim_{x \rightarrow \infty } \frac { \ln { x } }{ x } = 0 \qquad y = 0

Vertical asymptotes.

\lim_{x \rightarrow { 0 }^{ + } } \frac { \ln { x } }{ x } = \infty \qquad x = 0

 

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.