Solved Problems on Limits x and Continuity

1 Solved Problems on Limits and Continuity Mika Seppälä: Limits and Continuity Calculators Overview of Problems ( ) 2 0 sin lim x sin x → x x 1 2 2 3 ...

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Overview of Problems 1

Solved Problems on Limits and Continuity

3

5

lim

x 2 − 3x + 2 x →2 x −2

2

lim x 2 + 1 − x 2 − 1

4

x →∞

2x

lim

2x 2 + x + 1 − x 2 − 3 x + 1

x →0

7

lim

sin ( sin ( x ) )

x →0

9

lim

x →0

11 12 13

14

15

1

 1  Where f (φ ) = sin  2  is continuous?  φ − 1 x2 − x How must f ( 0 ) be determined so that f ( x ) = , x ≠ 0, x −1 is continuous at x = 0?

2

Which of the following functions have removable singularities at the indicated points?

3

x − 2x − 8 , x0 = −2, b) x+2

a)

f (x) =

c)

 1 h ( t ) = t sin   , t0 = 0 t 

g( x ) =

Mika Seppälä: Limits and Continuity

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lim

x →0

lim

sin ( 3 x )

x →0

6x

( )

sin x 2

x sin ( x )

x + 2sin ( x )

x 2 + 2 sin ( x ) + 1 − sin2 ( x ) − x + 1

10

lim e

tan ( x )

π

x→ + 2

Mika Seppälä: Limits and Continuity

The following undefined quantities cause problems: 0 ∞ 00 , , , ∞ − ∞, 0∞ , ∞0. 0 ∞ In the evaluation of expressions, use the rules a ∞ = 0, = ∞, ∞ × (negative number ) = −∞. ∞ positive number

x −1 , x0 = 1 x −1

Show that the equation sin ( x ) = e x has ∞ many solutions.

Calculators

lim x 2 + x + 1 − x 2 − x − 1

x →∞

Main Methods of Limit Computations

Where y = tan ( x ) is continuous?

2

x3 + x2 + x + 1 x3 + 3x2 + 5x + 2

x

Calculators

Overview of Problems

8

lim

x →∞

4

If the function, for which the limit needs to be computed, is defined by an algebraic expression, which takes a finite value at the limit point, then this finite value is the limit value. If the function, for which the limit needs to be computed, cannot be evaluated at the limit point (i.e. the value is an undefined expression like in (1)), then find a rewriting of the function to a form which can be evaluated at the limit point.

Calculators

Mika Seppälä: Limits and Continuity

1

Main Computation Methods 1 2

3

Frequently needed rule

Continuity of Functions

−b . 2

1

Cancel out common factors of rational functions. x 2 − 1 ( x − 1) ( x + 1) = = x + 1  → 2. x →1 x −1 x −1 If a square root appears in the expression, then multiply and divide by the conjugate of the square root expression. x +1 −

( x −2 = =

4

( a + b) ( a − b) = a

2

x +1 −

x +2

)(

x +1 +

( x + 1) − ( x − 2 ) x +1 +

Use the fact that lim x →0

Calculators

x −2

sin ( x ) x

=

x +1 +

x −2

)

x +1 +

x +2

 →0 x →∞

= 1.

Mika Seppälä: Limits and Continuity

Limits by Rewriting x − 3x + 2 x −2

A function f is continuous at a point x = a if lim f ( x ) = f ( a) . x →a

3

x −2 3

2

Functions defined by algebraic or elementary expressions involving polynomials, rational functions, trigonometric functions, exponential functions or their inverses are continuous at points where they take a finite well defined value.

4

Intermediate Value Theorem for Continuous Functions If f is continuous, f(a) < 0 and f(b) > 0, then there is a point c between a and b so that f(c) = 0.

Calculators

Mika Seppälä: Limits and Continuity

Limits by Rewriting

2

Problem 1

Solution

lim

x →2

Rewrite

x 2 − 3 x + 2 ( x − 1)( x − 2 ) = = x − 1. x −2 x −2

Hence lim

x →2

Calculators

x 2 − 3x + 2 = lim ( x − 1) = 1. x →2 x−2

Mika Seppälä: Limits and Continuity

Used to show that equations

The following are not continuous x = 0: have solutions. 1 x 1 f ( x ) = , g ( x ) = sin   ,h ( x ) = . x x x

Problem 2

lim

x →∞

x3 + x2 + x + 1 x3 + 3x 2 + 5x + 2

Solution

1 1+ + x3 + x2 + x + 1 x = x 3 + 3x 2 + 5x + 2 1+ 3 + x

Calculators

1 1 + x 2 x 3 →1. x →∞ 5 2 + 3 2 x x

Mika Seppälä: Limits and Continuity

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Limits by Rewriting lim x 2 + 1 − x 2 − 1

Problem 3

) ( 2

x2 + 1 −

(

)(

)

x2 + 1 − x2 − 1

x2 + 1 + x2 − 1

(

x + 1 + x2 − 1

)

2

(x

=

x +1+ x −1 2

2

) (

)=

+ 1 − x2 − 1

x +1+ x −1 2

2

2

x →∞

Calculators

x +1+ x −1

= 0.

2x 2 + x + 1 − x 2 − 3 x + 1

x →0

Rewrite

2x 2x

(

(

2

)(

2

Calculators

2

2

x+4

)

Rewrite

2x 2 + x + 1 + x 2 − 3 x + 1

) = 2x ( ( 2x + x + 1) − ( x − 3 x + 1) 2 ( 2 x + x + 1 + x − 3 x + 1) =  →1 2

)

2x + x + 1 + x − 3 x + 1 2

x →0

)

6x

Use the fact that lim

α →0

sin ( 3 x )

2x 2 + x + 1 + x 2 − 3 x + 1 x2 + 4x

sin ( 3 x )

lim

Solution

2x 2 + x + 1 + x 2 − 3 x + 1

2

2x + 2

Mika Seppälä: Limits and Continuity

Problem 6

=

2x 2 + x + 1 − x 2 − 3 x + 1

2

x2 + x + 1 + x2 − x − 1

Limits by Rewriting

2x 2 + x + 1 − x 2 − 3x + 1

2x

)=

2x

lim

(

) (

x2 + x + 1 + x2 − x − 1

+ x + 1 − x2 − x − 1

Calculators

Mika Seppälä: Limits and Continuity

Limits by Rewriting Solution

2

)

x2 + x + 1 − x2 − x − 1

x2 + x + 1 + x2 − x −1 x2 + x + 1 + x2 − x − 1 2 2+ Next divide by x. x =  →2 x →∞ 1 1 1 1 1+ + 2 + 1 − − 2 x x x x

2

x2 + 1 + x2 − 1

x →∞

(x =

2 2

2

Hence lim x + 1 − x − 1 = lim 2

Problem 5

Rewrite

x2 + x + 1 − x2 − x − 1 =

2

x2 − 1

2

=

x →∞

Solution

Rewrite

x2 + 1 − x2 − 1 =

(

lim x 2 + x + 1 − x 2 − x − 1

Problem 4

x →∞

Solution

=

Limits by Rewriting

Since lim

6x

=

sin ( 3 x )

x →0

3x

1 sin ( 3 x ) 2 3x

sin (α )

α

= 1.

= 1, we conclude that lim

x →0

sin ( 3 x ) 6x

=

1 . 2

Next divide by x.

x →0

Mika Seppälä: Limits and Continuity

Calculators

Mika Seppälä: Limits and Continuity

3

Limits by Rewriting Problem 7

lim

Limits by Rewriting

sin ( sin ( x ) )

x →0

lim

Solution

Rewrite:

x

Rewrite:

Solution

sin ( sin ( x ) )

=

x since lim

sin ( sin ( x ) ) sin ( x )

sin (α )

α

α →0

sin ( x )

x

( )

sin x 2

Problem 8

x →0

x sin ( x )

( ) = sin ( x )

sin x 2

 →1 x →0

2

x sin ( x )

x2

x  →1 sin ( x ) x →0

= 1. In the above, that fact

was applied first by substituting α = sin ( x ) . Hence lim

sin ( sin ( x ) ) sin ( x )

x →0

Calculators

= 1.

Calculators

Mika Seppälä: Limits and Continuity

Limits by Rewriting Problem 9

lim

x →0

Rewrite

Solution

x 2 + 2 sin ( x ) + 1 − sin2 ( x ) − x + 1

x 2 + 2 sin ( x ) + 1 − sin2 ( x ) − x + 1

( x + 2 sin ( x ) ) ( =

=

( x + 2 sin ( x )) (

Calculators

)

2

)(

x 2 + 2 sin ( x ) + 1 + sin2 ( x ) − x + 1

)

x 2 + 2 sin ( x ) + 1 + sin2 ( x ) − x + 1

) (

)

+ 2 sin ( x ) + 1 − sin2 ( x ) − x + 1

)

)

=

Mika Seppälä: Limits and Continuity

lim

x →0

x + 2 sin ( x )

x 2 + 2 sin ( x ) + 1 − sin2 ( x ) − x + 1

Rewrite x + 2 sin ( x )

( x + 2 sin ( x ) ) (

x 2 + 2sin ( x ) + 1 + sin2 ( x ) − x + 1

x − sin2 ( x ) + 2 sin ( x ) + x

)

Next divide by x.

2

 sin ( x )  1 + 2  x   =

(

x 2 + 2sin ( x ) + 1 + sin2 ( x ) − x + 1

sin ( x ) sin ( x ) x − sin ( x ) +2 +1 x x

x 2 + 2 sin ( x ) + 1 + sin2 ( x ) − x + 1

x 2 − sin2 ( x ) + 2 sin ( x ) + x

Problem 9

x 2 + 2 sin ( x ) + 1 − sin2 ( x ) − x + 1

x + 2 sin ( x ) + 1 + sin ( x ) − x + 1 2

( x + 2 sin ( x )) ( 2

Solution (cont’d)

=

x 2 + 2 sin ( x ) + 1 − sin2 ( x ) − x + 1

(x

Limits by Rewriting

x + 2 sin ( x )

x + 2 sin ( x )

(

Mika Seppälä: Limits and Continuity

)

 → x →0

3×2 = 2. 2 +1

Here we used the fact that all sin(x)/x terms approach 1 as x → 0. Calculators

Mika Seppälä: Limits and Continuity

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One-sided Limits lim e

Problem 10

Continuity

tan( x )

Problem 11

π

x→ + 2

Where the function y = tan ( x ) is continuous?

Solution Solution

For

π 2

The function y = tan ( x ) =

< x < π , tan ( x ) < 0 and lim tan ( x ) = −∞. π

x→ + 2

Hence lim e

tan( x )

π

x→ + 2

Calculators

Solution

Mika Seppälä: Limits and Continuity

Hence y = tan ( x ) is continuous at x ≠

π 2

+ nπ , n ∈ .

Calculators

Mika Seppälä: Limits and Continuity

Continuity

 1  Where the function f (φ ) = sin  2  is continuous?  φ − 1

 1  The function f (φ ) = sin  2  is continuous at all points  φ − 1 where it takes finite values.

 1  1 If φ = ±1, 2 is not finite, and sin  2  is undefined. φ −1  φ − 1  1  1 If φ ≠ ±1, 2 is finite, and sin  2  is defined and also finite. φ −1  φ − 1  1  Hence sin  2  is continuous for φ ≠ ±1.  φ − 1

Calculators

is continuous whenever cos ( x ) ≠ 0.

= 0.

Continuity Problem 12

sin ( x )

cos ( x )

Mika Seppälä: Limits and Continuity

Problem 13

How must f ( 0 ) be determined so that the function f (x) =

x2 − x , x ≠ 0, is continuous at x = 0 ? x −1

Solution

Condition for continuity of a function f at a point x0 is:

lim f ( x ) = f ( x0 ) . Hence f ( 0 ) must satisfy f ( 0 ) = lim f ( x ) .

x → x0

x →0

x ( x − 1) x2 − x = lim = lim x = 0. x →0 x − 1 x →0 x →0 x −1

Hence f ( 0 ) = lim

Calculators

Mika Seppälä: Limits and Continuity

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Continuity

Continuity

A number x0 for which an expression f ( x ) either is undefined or

Problem 15

infinite is called a singularity of the function f . The singularity is

inifinitely many solutions.

said to be removable, if f ( x0 ) can be defined in such a way that Solution

the function f becomes continous at x = x0 . Problem 14

Which of the following functions have removable singularities at the indicated points?

a) b) c)

Calculators

x 2 − 2x − 8 f (x) = , x 0 = −2 x +2 x −1 g( x ) = , x0 = 1 x −1  1 h ( t ) = t sin   , t 0 = 0 t 

Mika Seppälä: Limits and Continuity

Answer Removable Not removable Removable

Show that the equation sin ( x ) = e x has

sin ( x ) = e x ⇔ f ( x ) = sin ( x ) − e x = 0.

By the intermediate Value Theorem, a continuous function takes any value between any two of its values. I.e. it suffices to show that the function f changes its sign infinitely often. n π  Observe that 0 < e x < 1 for x < 0, and that sin  + nπ  = ( −1) , n ∈ . 2  π Hence f ( x ) < 0 for x = + nπ if n is an odd negative number 2 π and f ( x ) > 0 for x = + nπ if n is an even negative number. 2 π π  We conclude that every interval  + 2nπ , + ( 2n + 1) π  , n ∈ and n < 0, contains 2 2  a solution of the original equation. Hence there are infinitely many solutions. Calculators

Mika Seppälä: Limits and Continuity

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