SOLUTIONS TO CONTINUITY OF FUNCTIONS OF ONE VARIABLE

SOLUTION 1 : Function f is defined at x=1 since

i.) f(1) = 2 .

The limit

$${ \lim_{ x \to 1 } f(x) = \lim_{ x \to 1 } (3x-5) }$$

= 3 (1) - 5

= -2 ,

i.e.,

ii.) $${ \lim_{ x \to 1 } f(x) = -2 }$$ .

But

iii.) $${ \lim_{ x \to 1 } f(x) \ne f(1) }$$ ,

so condition iii.) is not satisfied and function f is NOT continuous at x=1 .

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SOLUTION 2 : Function f is defined at x=-2 since

i.) f(-2) = (-2)² + 2(-2) = 4-4 = 0 .

The left-hand limit

$${ \lim_{ x \to -2^{-} } f(x) = \lim_{ x \to -2^{-} } (x^2 + 2x) }$$

= (-2)² + 2(-2)

= 4 - 4

= 0 .

The right-hand limit

$${ \lim_{ x \to -2^{+} } f(x) = \lim_{ x \to -2^{+} } (x^3 - 6x) }$$

= (-2)³ - 6(-2)

= -8 + 12

= 4 .

Since the left- and right-hand limits are not equal, ,

ii.) $${ \lim_{ x \to -2 } f(x) }$$ does not exist,

and condition ii.) is not satisfied. Thus, function f is NOT continuous at x=-2 .

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SOLUTION 3 : Function f is defined at x=0 since

i.) f(0) = 2 .

The left-hand limit

$${ \lim_{ x \to 0^{-} } f(x) = \lim_{ x \to 0^{-} } { x-6 \over x-3 } }$$

$= \displaystyle{ -6 \over -3 }$

= 2 .

The right-hand limit

$${ \lim_{ x \to 0^{+} } f(x) = \lim_{ x \to 0^{+} } \sqrt{ 4 + x^2 } }$$

$= \sqrt{ 4 + (0)^2 }$

$= \sqrt{ 4 }$

= 2 .

Thus, $${ \lim_{ x \to 0 } f(x) }$$ exists with

ii.) $${ \lim_{ x \to 0 } f(x) = 2 }$$ .

Since

iii.) $${ \lim_{ x \to 0 } f(x) = 2 = f(0) }$$ ,

all three conditions are satisfied, and f is continuous at x=0 .

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SOLUTION 4 : Function h is not defined at x=-1 since it leads to division by zero. Thus,

i.) h(-1)

does not exist, condition i.) is violated, and function h is NOT continuous at x = -1 .

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SOLUTION 5 : First, check for continuity at x=3 . Function f is defined at x=3 since

i.) $f(3) = \displaystyle{ 9 \over 2 }$ .

The limit

$${ \lim_{ x \to 3 } f(x) = \lim_{ x \to 3 } \ { x^3-27 \over x^2-9 } } = \displaystyle{ ^{^{^{^{''}}}}{0 \over 0}^{''} }$$

(Circumvent this indeterminate form by factoring the numerator and the denominator.)

$${ = \lim_{ x \to 3 } \ { x^3- 3^3 \over x^2-3^2 } }$$

(Recall that A² - B² = (A-B)(A+B) and A³ - B³ = (A-B)(A²+AB+B² ) . )

$${ = \lim_{ x \to 3 } { (x-3)(x^2+3x+9) \over (x-3)(x+3) } }$$

(Divide out a factor of (x-3) . )

$${ = \lim_{ x \to 3 } { x^2+3x+9 \over x+3 } }$$

=$${ (3)^2 + 3(3) + 9 \over (3) + 3 }$$

$= \displaystyle{ 27 \over 6 }$

$= \displaystyle{ 9 \over 2 }$ ,

i.e.,

ii.) $${ \lim_{ x \to 3 } f(x) = \displaystyle{ 9 \over 2 } }$$ .

Since,

iii.) $${ \lim_{ x \to 3 } f(x) = \displaystyle{ 9 \over 2 } = f(3) }$$ ,

all three conditions are satisfied, and f is continuous at x=3 . Now, check for continuity at x=-3 . Function f is not defined at x = -3 because of division by zero. Thus,

i.) f(-3)

does not exist, condition i.) is violated, and f is NOT continuous at x=-3 .

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SOLUTION 6 : Functions y = x² + 3x + 5 and y = x² + 3x - 4 are continuous for all values of x since both are polynomials. Thus, the quotient of these two functions, $f(x) = \displaystyle{ x^2 + 3x + 5 \over x^2 + 3x - 4 }$ , is continuous for all values of x where the denominator, y = x² + 3x - 4 = (x-1)(x+4) , does NOT equal zero. Since (x-1)(x+4) = 0 for x=1 and x=-4 , function f is continuous for all values of x EXCEPT x=1 and x=-4 .

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SOLUTION 7 : First describe function g using functional composition. Let f(x) = x^1/3 , $h(x) = \sin x$ , and k(x) = x²⁰ + 5 . Function k is continuous for all values of x since it is a polynomial, and functions f and h are well-known to be continuous for all values of x . Thus, the functional compositions

$h( k(x) ) = \sin( k(x) ) = \sin(x^{20} + 5 )$

and

$f( h ( k(x) ) ) = ( h( k(x) ) )^{1/3} = ( \sin(x^{20} + 5) )^{1/3}$

are continuous for all values of x . Since

$g(x) = ( \sin(x^{20} + 5) )^{1/3} = f( h ( k(x) ) )$ ,

function g is continuous for all values of x .

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SOLUTION 8 : First describe function f using functional composition. Let g(x) = x² - 2x and $h(x) = \sqrt{ x }$ . Function g is continuous for all values of x since it is a polynomial, and function h is well-known to be continuous for $x \ge 0$ . Since g(x) = x² - 2x = x(x-2) , it follows easily that $g(x) \ge 0$ for $x \le 0$ and $x \ge 2$ . Thus, the functional composition

$h( g(x) ) = \sqrt{ g(x) } = \sqrt{ x^2 - 2x }$

is continuous for $x \le 0$ and $x \ge 2$ . Since

$f(x) = \sqrt{ x^2 - 2x } = h( g(x) )$ ,

function f is continuous for $x \le 0$ and $x \ge 2$ .

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SOLUTION 9 : First describe function f using functional composition. Let $g(x) = \displaystyle{ x-1 \over x+2 }$ and $h(x) = \ln{ x }$ . Since g is the quotient of polynomials y = x-1 and y = x+2 , function g is continuous for all values of x EXCEPT where x+2 = 0 , i.e., EXCEPT for x = -2 . Function h is well-known to be continuous for x > 0 . Since $g(x) = \displaystyle{ x-1 \over x+2 }$ , it follows easily that g(x) > 0 for x < -2 and x > 1 . Thus, the functional composition

$h( g(x) ) = \ln{ (g(x) ) } = \ln{ \Big( \displaystyle{ x-1 \over x+2 } \Big) }$

is continuous for x < -2 and x > 1 . Since

$f(x) = \ln{ \Big( \displaystyle{ x-1 \over x+2 } \Big) }= h( g(x) )$ ,

function f is continuous for x < -2 and x > 1 .

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SOLUTION 10 : First describe function f using functional composition. Let $g(x) = \sin x$ and h(x) = e ^x , both of which are well-known to be continuous for all values of x . Thus, the numerator $y = e^{ \sin x } = h ( g(x) )$ is continuous (the functional composition of continuous functions) for all values of x . Now consider the denominator $y = 4 - \sqrt{ x^2 - 9 }$ . Let g(x) = 4 , h(x) = x² - 9 , and $k(x) = \sqrt{ x }$ . Functions g and h are continuous for all values of x since both are polynomials, and it is well-known that function k is continuous for $x \ge 0$ . Since h(x) = x² - 9 = (x-3)(x+3) = 0 when x=3 or x=-3 , it follows easily that $h(x) \ge 0$ for $x \ge 3$ and $x \le -3$ , so that $y = \sqrt{ x^2 - 9 } = k ( h(x) )$ is continuous (the functional composition of continuous functions) for $x \ge 3$ and $x \le -3$ . Thus, the denominator $y = 4 - \sqrt{ x^2 - 9 }$ is continuous (the difference of continuous functions) for $x \ge 3$ and $x \le -3$ . There is one other important consideration. We must insure that the DENOMINATOR IS NEVER ZERO. If

$y = 4 - \sqrt{ x^2 - 9 } = 0$

then

$4 = \sqrt{ x^2 - 9 }$ .

Squaring both sides, we get

16 = x² - 9

so that

x² = 25

when

x = 5 or x = -5 .

Thus, the denominator is zero if x = 5 or x = -5 . Summarizing, the quotient of these continuous functions, $f(x) = \displaystyle{ e^{ \sin x } \over 4 - \sqrt{ x^2 - 9 } }$ , is continuous for $x \ge 3$ and $x \le -3$ , but NOT for x = 5 and x = -5 .

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SOLUTION 11 : Consider separately the three component functions which determine f . Function $y = \displaystyle{ x-1 \over \sqrt{ x } - 1 }$ is continuous for x > 1 since it is the quotient of continuous functions and the denominator is never zero. Function y = 5 -3x is continuous for $-2 \le x \le 1$ since it is a polynomial. Function $y = \displaystyle{ 6 \over x-4 }$ is continuous for x < -2 since it is the quotient of continuous functions and the denominator is never zero. Now check for continuity of f where the three components are joined together, i.e., check for continuity at x=1 and x=-2 . For x = 1 function f is defined since

i.) f(1) = 5 - 3(1) = 2 .

The right-hand limit

$${ \lim_{ x \to 1^{+} } f(x) }$$ = $${ \lim_{ x \to 1^{+} } { x-1 \over \sqrt{ x } - 1 } } = \displaystyle{ ^{^{^{^{''}}}}{0 \over 0}^{''} }$$

(Circumvent this indeterminate form one of two ways. Either factor the numerator as the difference of squares, or multiply by the conjugate of the denominator over itself.)

$= \displaystyle{ \lim_{ x \to 1^{+} } { (\sqrt{ x })^2 - (1)^2 \over \sqrt{ x } - 1 } }$

$= \displaystyle{ \lim_{ x \to 1^{+} } { (\sqrt{ x } - 1) (\sqrt{ x } + 1) \over \sqrt{ x } - 1 } }$

$= \displaystyle{ \lim_{ x \to 1^{+} } (\sqrt{ x } + 1) }$

$= (\sqrt{ 1 } + 1)$

= 2 .

The left-hand limit

$${ \lim_{ x \to 1^{-} } f(x) }$$ = $${ \lim_{ x \to 1^{-} } (5-3x) }$$

= 5 - 3(1)

= 2 .

Thus,

ii.) $${ \lim_{ x \to 1 } f(x) } = 2$$ .

Since

iii.) $${ \lim_{ x \to 1 } f(x) } = 2 = f(1)$$ ,

all three conditions are satisfied, and function f is continuous at x=1 . Now check for continuity at x=-2 . Function f is defined at x=-2 since

i.) f(-2) = 5 - 3(-2) = 11 .

The right-hand limit

$${ \lim_{ x \to -2^{+} } f(x) }$$ = $${ \lim_{ x \to -2^{+} } (5-3x) }$$

= 5 - 3( -2)

= 11 .

The left-hand limit

$${ \lim_{ x \to -2^{-} } f(x) }$$ = $${ \lim_{ x \to -2^{-} } { 6 \over x-4 } }$$

$= \displaystyle{ 6 \over (-2) -4 }$

$= \displaystyle{ 6 \over -6 }$

= -1 .

Since the left- and right-hand limits are different,

ii.) $${ \lim_{ x \to -2 } f(x) }$$ does NOT exist,

condition ii.) is violated, and function f is NOT continuous at x=-2 . Summarizing, function f is continuous for all values of x EXCEPT x=-2 .

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SOLUTION 12 : First, consider separately the two components which determine function f . Function y = A² x - A is continuous for $x \ge 3$ for any value of A since it is a polynomial. Function y = 4 is continuous for x < 3 since it is a polynomial. Now determine A so that function f is continuous at x=3 . Function f must be defined at x=3 , so

i.) f(3)= A² (3) - A = 3 A² - A .

The right-hand limit

$${ \lim_{ x \to 3^{+} } f(x) }$$ = $${ \lim_{ x \to 3^{+} } (A^2 x - A) }$$

= A² (3) - A

= 3 A² - A .

The left-hand limit

$${ \lim_{ x \to 3^{-} } f(x) }$$ = $${ \lim_{ x \to 3^{-} } 4 }$$

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) $${ \lim_{ x \to 3 } f(x) } = 3 A^2 - A = 4$$ ,

so that

3 A² - A - 4 = 0 .

Factoring, we get

(3A - 4)(A + 1) = 0

for

$A = \displaystyle{ 4 \over 3 }$ or A = -1 .

For either choice of A ,

iii.) $${ \lim_{ x \to 3 } f(x) } = 4 = f(3)$$ ,

all three conditions are satisfied, and f is continuous at x=3 . Therefore, function f is continuous for all values of x if $A = \displaystyle{ 4 \over 3 }$ or A = -1 .

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SOLUTION 13 : First, consider separately the three components which determine function f . Function y = Ax - B is continuous for $x \le -1$ for any values of A and B since it is a polynomial. Function y = 2x² + 3Ax + B is continuous for $-1 < x \le 1$ for any values of A and B since it is a polynomial. Function y = 4 is continuous for x > 1 since it is a polynomial. Now determine A and B so that function f is continuous at x=-1 and x=1 . First, consider continuity at x=-1 . Function f must be defined at x=-1 , so

i.) f(-1)= A(-1) - B = - A - B .

The left-hand limit

$${ \lim_{ x \to -1^{-} } f(x) }$$ = $${ \lim_{ x \to -1^{-} } (Ax - B) }$$

= A (-1) - B

= - A - B .

The right-hand limit

$${ \lim_{ x \to -1^{+} } f(x) }$$ = $${ \lim_{ x \to -1^{+} } (2x^2 + 3Ax + B) }$$

= 2(-1)² + 3A(-1) + B

= 2 - 3A + B .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) $${ \lim_{ x \to -1 } f(x) } = - A - B = 2 - 3A + B$$ ,

so that

2A - 2B = 2 ,

or

(Equation 1)

A - B = 1 .

Now consider continuity at x=1 . Function f must be defined at x=1 , so

i.) f(1)= 2(1)² + 3A(1) + B = 2 + 3A + B .

The left-hand limit

$${ \lim_{ x \to 1^{-} } f(x) }$$ = $${ \lim_{ x \to 1^{-} } (2x^2 + 3Ax + B ) }$$

= 2(1)² + 3A(1) + B

= 2 + 3A + B .

The right-hand limit

$${ \lim_{ x \to 1^{+} } f(x) }$$ = $${ \lim_{ x \to 1^{+} } 4 }$$

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) $${ \lim_{ x \to 1 } f(x) } = 2 + 3A + B = 4$$ ,

or

(Equation 2)

3A + B = 2 .

Now solve Equations 1 and 2 simultaneously. Thus,

A - B = 1 and 3A + B = 2

are equivalent to

A = B + 1 and 3A + B = 2 .

Use the first equation to substitute into the second, getting

3 (B + 1 ) + B = 2 ,

3 B + 3 + B = 2 ,

and

4 B = -1 .

Thus,

$B = \displaystyle{ -1 \over 4 }$

and

$A = B + 1 = \displaystyle{ -1 \over 4 } + 1 = \displaystyle{ 3 \over 4 }$ .

For this choice of A and B it can easily be shown that

iii.) $${ \lim_{ x \to 1 } f(x) } = 4 = f(1)$$

and

iii.) $${ \lim_{ x \to -1 } f(x) } = \displaystyle{ -1 \over 2 } = f(-1)$$ ,

so that all three conditions are satisfied at both x=1 and x=-1 , and function f is continuous at both x=1 and x=-1 . Therefore, function f is continuous for all values of x if $A = \displaystyle{ 3 \over 4 }$ and $B = \displaystyle{ -1 \over 4 }$ .

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SOLUTION 14 : First describe f using functional composition. Let g(x) = -1/x² and h(x) = e^x . Function h is well-known to be continuous for all values of x . Function g is the quotient of functions continuous for all values of x , and is therefore continuous for all values of x except x=0 , that x which makes the denominator zero. Thus, for all values of x except x=0 ,

f(x) = h ( g(x) ) = e ^g(x) = e ^-1/x2

is a continuous function (the functional composition of continuous functions). Now check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .

The limit

$${ \lim_{ x \to 0 } { -1 \over x^2 } = \displaystyle{ ^{^{^{^{''}}}}{-1 \over 0^{+} } ^{''} } }$$

(The numerator approaches -1 and the denominator is a positive number approaching zero.)

$= - \infty$ ,

so that

$${ \lim_{ x \to 0 } { f(x) } } = \displaystyle{ \lim_{ x \to 0 } e^{ -1 / x^2 } }$$

$= \displaystyle{ ^{'' } { e^{ - \infty } }^{''} }$

$= \displaystyle{ ^{^{^{^{''}}}}{ 1 \over e^{ \infty } }^{''} }$

$= \displaystyle{ ^{^{^{^{''}}}}{1 \over { \infty } }^{''} }$

= 0 ,

i.e.,

ii.) $${ \lim_{ x \to 0 } { f(x) } } = 0$$ .

Since

iii.) $${ \lim_{ x \to 0 } { f(x) } } = 0 = f(0)$$ ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all values of x .

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SOLUTION 15 : First show that f is continuous for all values of x . Describe f using functional composition. Let $g(x) = \displaystyle{ 1 \over x }$ , $h(x) = \cos x$ , and k(x) = x² . Function h is well-known to be continuous for all values of x . Function k is a polynomial and is therefore continuous for all values of x . Function g is the quotient of functions continuous for all values of x , and is therefore continuous for all values of x except x=0 , that x which makes the denominator zero. Thus, for all values of x except x=0 ,

$f(x) = k(x) h( g(x) ) = x^2 \cos( g(x) ) = x^2 \cos \Big( \displaystyle{ 1 \over x } \Big)$

is a continuous function (the product and functional composition of continuous functions). Now check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .

The limit $${ \lim_{ x \to 0 } \cos \Big( { 1 \over x } \Big) }$$ does not exist since the values of $\cos \Big( \displaystyle{ 1 \over x } \Big)$ oscillate between -1 and +1 as x approaches zero. However, for $x \ne 0$

$-1 \le \cos \Big( \displaystyle{ 1 \over x } \Big) \le +1$

so that

$-x^2 \le x^2 \cos \Big( \displaystyle{ 1 \over x } \Big) \le x^2$.

Since

$${ \lim_{ x \to 0 } { (-x^2 ) } } = 0 = \displaystyle{ \lim_{ x \to 0 } { x^2 } }$$ ,

it follows from the Squeeze Principle that

ii.) $${ \lim_{ x \to 0 } f(x) } = \displaystyle{ \lim_{ x \to 0 } x^2 \cos \Big( { 1 \over x } \Big) } = 0$$ .

Since

iii.) $${ \lim_{ x \to 0 } { f(x) } } = 0 = f(0)$$ ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all values of x . Now show that f is differentiable for all values of x . For $x \ne 0$ we can differentiate f using the product rule and the chain rule. That is, for $x \ne 0$ the derivative of f is

$f'(x) = x^2 \ D \Big\{ \cos \Big( \displaystyle{ 1 \over x } \Big) \Big\} + D \{ x^2 \} \ \cos \Big( \displaystyle{ 1 \over x } \Big)$

$= x^2 \Big\{-\sin \Big( \displaystyle{ 1 \over x } \Big) \ D \Big\{ \displayst... ...\over x } \Big\} \Big\} + \{ 2x \} \cos \Big( \displaystyle{ 1 \over x } \Big)$

$= - x^2 \sin \Big( \displaystyle{ 1 \over x } \Big) \Big\{ \displaystyle{ -1 \over x^2 } \Big\} + 2x \cos \Big( \displaystyle{ 1 \over x } \Big)$

$= \sin \Big( \displaystyle{ 1 \over x } \Big) + 2x \cos \Big( \displaystyle{ 1 \over x } \Big)$ .

Use the limit definition of the derivative to differentiate f at x=0 . Then

$f'(0) = \displaystyle{ \lim_{ \Delta x \to 0 } { f(0 + \Delta x) - f(0) \over \Delta x } }$

$= \displaystyle {\lim_{\Delta x\to 0} } \; \;\displaystyle { {f( \Delta x) - 0 } \over {\Delta x} }$

$= \displaystyle {\lim_{\Delta x\to 0} } \; \;\displaystyle{ ( \Delta x)^2 \cos \Big( \displaystyle{ 1 \over \Delta x } \Big) \over {\Delta x} }$

$= \displaystyle {\lim_{\Delta x\to 0} } \; \;\displaystyle{ \Delta x \cos \Big( \displaystyle{ 1 \over \Delta x } \Big) }$ .

Use the Squeeze Principle to evaluate this limit. For $\Delta x \ne 0$

$-1 \le \cos \Big( \displaystyle{ 1 \over \Delta x } \Big) \le +1$ .

If $\Delta x > 0$ , then

$- \Delta x \le \Delta x \cos \Big( \displaystyle{ 1 \over \Delta x } \Big) \le \Delta x$ .

If $\Delta x < 0$ , then

$- \Delta x \ge \Delta x \cos \Big( \displaystyle{ 1 \over \Delta x } \Big) \ge \Delta x$ .

In either case,

$${\lim_{\Delta x\to 0} } \; \;( - \Delta x ) = 0 = \displaystyle {\lim_{\Delta x\to 0} } \; \;\Delta x$$ ,

and it follows from the Squeeze Principle that

$f'(0) = \displaystyle {\lim_{\Delta x\to 0} } \; \;\displaystyle{ \Delta x \cos \Big( \displaystyle{ 1 \over \Delta x } \Big) } = 0$ .

Thus, f is differentiable for all values of x . Check to see if f' is continuous at x=0 . The function f' is defined at x=0 since

i.) f'(0) = 0 .

However,

ii.) $${ \lim_{ x \to 0 } f'(x) } = \displaystyle{ \lim_{ x \to 0 } \B... ...e{ 1 \over x } \Big) + 2x \cos \Big( \displaystyle{ 1 \over x } \Big) \Big\} }$$

does not exist since the values of $\sin \Big( \displaystyle{ 1 \over x } \Big)$ oscillate between -1 and +1 as x approaches zero. Thus, condition ii.) is violated, and the derivative , f' , is not continuous at x=0 .

NOTE : The continuity of function f for all values of x also follows from the fact that f is differentiable for all values of x .

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