Exercise 1

Using the definition of a limit, prove that:

\lim_{ x \rightarrow 1 } \frac { x + 3 }{ 2 } = 2

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Exercise 2

Using the graph of the function f(x), determine the following limits.

Exercise 3

Using the definition of a limit, prove that:

f(x) = \left\{\begin{matrix} 2x - 1 \quad if \quad x < 0 \\ \frac { { x }^{ 2 } - 3  }{ 3 } \quad if \quad x > 0 \end{matrix}\right has a limit 1 as x 0

Calculate the Following Limits:

Exercise 4

\lim_{ x \rightarrow 3 } ( \frac { x - 1 }{ x - 3 } - \frac { x + 5 }{ { x }^{ 2 } - 4 x + 3 })

Exercise 5

\lim_{ x \rightarrow \infty } \frac { { 2 }^{ x + 1 } + { 3 }^{ x + 1 } }{ { 2 }^{ x } + { 3 }^{ x } }

Exercise 6

\lim_{ x \rightarrow 0 } \frac { \sqrt { x + 9 } - 3 }{ \sqrt { x + 16 } - 4 }

Exercise 7

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

 

Solution of exercise 1

Using the definition of a limit, prove that:

\lim_{ x \rightarrow 1 } \frac { x + 3 }{ 2 } = 2

\left | \frac { x + 3 }{ 2 } - 2 \right | < \epsilon \qquad \left | \frac { x + 3 - 4 }{ 2 } \right | < \epsilon \qquad \left | \frac { x - 1 }{ 2 } \right | < \epsilon

\frac { \left | x - 1 \right | }{ 2 } < \epsilon \qquad \left | x - 1 \right | < 2 \epsilon

If \delta = 2 \epsilon

\lim_{ x \rightarrow 1 } \frac { x + 3 }{ 2 } = 2 \Leftrightarrow \forall \epsilon > 0 \quad \exists  \delta ( \epsilon ) > 0 / 0 < \left | x - 1 \right | < 2 \epsilon \Rightarrow  \left | \frac { x + 3 }{ 2 } - 2 \right | < \epsilon

To check this, take a \epsilon = 0.01.

{ E }_{ 2 \epsilon } (1) = { E }_{ 0.02 } (1) = (1 - 0.02, 1 + 0.02) = (0.98, 2.02)

{ E }_{ \epsilon } (1) = { E }_{ 0.01 } (1) = (2 - 0.01, 2 + 0.01) = (0.99, 2.01)

For x = 0.995 \quad f(x) = \frac { 0.995 + 3 }{ 2 } = 1.9975.

For x = 1.015 \quad f(x) = \frac { 1.015 + 3 }{ 2 } = 2.0075.

 

Solution of exercise 2

Using the graph of the function f(x), determine the following limits.

  1. \lim_{ x \rightarrow - \infty } f(x) = - \infty
  2. \lim_{ x \rightarrow - 1 } f(x) = \left\{\begin{matrix} \lim_{ x \rightarrow { -1 }^{ - } } f(x) = -\infty \\ \lim_{ x \rightarrow { -1 }^{ + } } f(x) = \infty \end{matrix}\right
  3. \lim_{ x \rightarrow { 1 }^{ - } } f(x) = - \infty
  4. \lim_{ x \rightarrow { 1 }^{ + } } f(x) = \infty
  5. \lim_{ x \rightarrow \infty } f(x) = \infty

 

Solution of exercise 3

Using the definition of a limit, prove that:

f(x) = \left\{\begin{matrix} 2x - 1 \quad if \quad x < 0 \\ \frac { { x }^{ 2 } - 3  }{ 3 } \quad if \quad x > 0 \end{matrix}\right has a limit 1 as x 0

f(x) = \left\{\begin{matrix} 2x - 1 \quad if \quad x < 0 \\ \frac { { x }^{ 2 } - 3  }{ 3 } \quad if \quad x > 0 \end{matrix}\right

Left side limit.

\left | (2x - 1) - (-1) \right | < \epsilon \qquad \left | 2x \right | < \epsilon; 2 \left | x \right | < \epsilon

\left | x \right | < \frac { \epsilon }{ 2 } \righrarrow \left | x - 0 \right | < \frac { \epsilon }{ 2 } = \delta

if \quad \left | x - 0 \right | < \delta \rightarrow \left | f(x) - (-1) \right | < \epsilon

\lim_{ x \rightarrow { 0 }^{ - } } = -1

Right side limit

\left | \frac { { x }^{ 2 } - 3 }{ 3 } - (-1) \right | < \epsilon \qquad \left | \frac { { x }^{ 2 } }{ 3 } \right | < \epsilon \qquad \left | { x }^{ 2 } \right | < 2 \epsilon

\left | x \right | < \sqrt { 3 \epsilon } \rightarrow \left | x - 0 \right | < \sqrt { 3 \epsilon } = \delta

If \quad \left | x - 0 \right | < \delta \rightarrow \left | f(x) - (-1) \right | < \epsilon

\lim_{ x \rightarorw { 0 }^{ + } } f(x) = -1

\lim_{ x \rightarrow { 0 }^{ - } } f(x) = -1

Solution of exercise 4

Calculate the limit:

\lim_{ x \rightarrow 3 } ( \frac { x - 1 }{ x - 3 } - \frac { x + 5 }{ { x }^{ 2 } - 4 x + 3 })

\lim_{ x \rightarrow 3 } ( \frac { x - 1 }{ x - 3 } - \frac { x + 5 }{ { x }^{ 2 } - 4 x + 3 }) = \frac { \infty }{ \infty }

\lim_{ x \rightarrow 3 } ( \frac { x - 1 }{ x - 3 } - \frac { x + 5 }{ { x }^{ 2 } - 4 x + 3 }) = \lim_{ x \rightarrow 3 } \frac { ({ x }^{ 2 } - 2x + 1) - (x + 5) }{ (x - 3)(x - 1)} = \frac { -4 }{ 0 }

Calculate the side limits to determine the sign of \infty.

\lim_{ x \rightarrow { 3 }^{ - } } \frac { { x }^{ 2 } - 3x - 4 }{ (x - 3)( x - 1) } = \infty

\lim_{ x \rightarrow { 3 }^{ + } } \frac { { x }^{ 2 } - 3x - 4 }{ (x - 3)( x - 1) } = - \infty

No limit.

 

Solution of exercise 5

Calculate the limit:

\lim_{ x \rightarrow \infty } \frac { { 2 }^{ x + 1 } + { 3 }^{ x + 1 } }{ { 2 }^{ x } + { 3 }^{ x } }

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

\lim_{ x \rightarrow \infty } \frac { { 2 }^{ x } . 2 + { 3 }^{ x } . 3 }{ { 2 }^{ x } + { 3 }^{ x } } = \lim_{ x \rightarrow \infty } \frac { 2 . { (\frac { 2 }{ 3 }) }^{ x } + 3 }{ { (\frac { 2 }{ 3 }) }^{ x } + 1 } = 3

 

Solution of exercise 6

Calculate the limit:

\lim_{ x \rightarrow 0 } \frac { \sqrt { x + 9 } - 3 }{ \sqrt { x + 16 } - 4 }

\lim_{ x \rightarrow 0 } \frac { \sqrt { x + 9 } - 3 }{ \sqrt { x + 16 } - 4 } = \frac { 0 }{ 0 }

\lim_{ x \rightarrow 0 } \frac { (\sqrt { x + 9 } - 3)( \sqrt { x + 9 } + 3)( \sqrt { x + 16 } + 4) }{ (\sqrt { x + 16 } - 4)( \sqrt { x + 16 } + 4)( \sqrt { x + 9 } + 3) }

\lim_{ x \rightarrow 0 } \frac { (x + 9 - 9 )( \sqrt { x + 16 } + 4 )}{ ( x + 16 - 16 )( \sqrt { x + 9 } + 3 )}

\lim_{ x \rightarrow 0 } \frac { (\sqrt { x + 16 } + 4) }{ (\sqrt { x + 9} + 3) } = \frac { 8 }{ 6 } = \frac { 4 }{ 3 }

 

Solution of exercise 7

Calculate the limit:

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

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

\lim_{ x \rightarrow \infty } { (1 + \frac { -3 }{ 2x + 4 }) }^{ \frac { { x }^{ 2 } }{ x + 1 } } = \lim_{ x \rightarrow \infty } { (1 + \frac { 1 }{ \frac { 2x + 4 }{ -3 } }) }^{ \frac { { x }^{ 2 } }{ x + 1 } }

{ [ \lim_{ x \rightarrow \infty } { (1 + \frac { 1 }{ \frac { 2x + 4 }{ -3 } }) }^{ \frac { 2x + 4 }{ -3 } }] } ^{ \lim_{ x \rightarrow \infty } (\frac { -3 }{ 2x + 4 } . \frac { { x }^{ 2 } }{ x + 1 } )  } = { e }^{ \frac { -3 }{ 2 } } = \frac { 1 }{ \sqrt { { e }^{ 3 } }}

 

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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.