LIMITS e t −1
The lim is 1.
x →0 t
Logarithmic Function
The logarithmic form of x=b y is log b x= y .
lim log b x is +∞
Given that b>0, x→+∞
lim ¿ is -∞ CONTINUITY
Given that b>0, +¿
x→ 0 log b x ¿
lim ln x is +∞ A function f (x) is continuous at a number
x→+∞ c if all of the following conditions are
lim ¿ satisfied:
a. f (c ) exists
+¿
x→ 0 ln 4 xis−∞ ¿
A vertical asymptote is a vertical line that
b. lim f (x )exists
the graph of the function approaches as f x →c
(x) approaches positive infinity or negative c. f ( c ) =lim
x→ c
f ( x)
infinity.
Note: Logarithmic functions have vertical DISCONTINUITY
asymptotes in which it is the restriction of
the function. If one of the three conditions was not
satisfied, then the function is
Exponential Function discontinuous.
Types of Discontinuity:
Functions having an exponent that is/has a. Removable Discontinuity- it occurs
a variable and whose base is greater than when there is a hole in the graph of
zero and is not equal to 1 is called a function.
exponential function. Note: We can redefine the function
x
lim b is + ∞
Given that b>0, x→+∞ to remove the discontinuity.
lim b is 0 x 2 x2 +5 x−3
Given that b>0, x→−∞ Example: f ( x )= , the redefined
x +3
x
lim b is 0
Given that 0<b<1, x→+∞ function is:
2 x 2 +5 x−3
lim e is +∞
The x→+∞
x
lim b x is + ∞
Given that 0<b<1, x→−∞
{
f ( x )= x+ 3
; x ≠−3
−7 ; x=−3
x
Steps: 1. Copy the original function then x ≠ should
lim e is 0
The x→−∞ not be equal to the restriction.
−x
lim e
The x→+∞ is 0. 2. Find the limit then x = restriction
−x
lim e
The x→−∞ is +∞ b. Jump Discontinuity- it occurs when
A horizontal asymptote is a horizontal line the graph of the function stops at one
that the graph of the function approaches point and seems to jump at another point.
as x approaches positive infinity or In a jump discontinuity, the left hand and
negative infinity. the right hand limits exist but are not
Note: An exponential function has a equal.
horizontal asymptote if the limit exists. c. Infinite Discontinuity- it occurs when
Whatever the limit of the function if it the function has at least one infinite limit.
exists is automatically the horizontal
asymptote of the function. Intermediate Value Theorem
Rational functions have either horizontal Intermediate Value theorem states that if the
or vertical asymptote and some have both. function f (x) is continuous on the closed interval
[a,b] from point A to point B, then the curve
Trigonometric Functions crosses every line y= y3 in at least one point and
sin t
The lim is 1. this line lies between the line y= y1 and y= y 2 .
x →0 t
1−cos t
The lim is 0.
x →0 t
In other words a function f (x) which is found to As Q approaches point P, then PQ will be
be continuous over a closed interval [a , b] will tangent to the graph.
The tangent line to y=f ( x )at point P is the
take any value between f (a) and f (b).
limiting position of all secant lines PQ as point
Extreme Value Theorem Q approaches point P.
The figure illustrates The Extreme Value Theorem Equation of the Tangent Line
which states that a function f (x) which is found to
be continuous over a closed interval [a,b] is The steps in finding the equation of the tangent
guaranteed to have extreme values in that line of any function are:
interval.
Find the value of ( x 0 , y 0 ).
An extreme value of f , or extremum, is either a Get the slope of the tangent line by
minimum or a maximum value of the function. A computing
minimum value of f occurs at some x=c if y− y 0
f (c )≤ f ( x ) for all x ≠ c in the interval. A maximum m= lim
x→ x0 x−x 0
value of f occurs at some x=c if f (c )≤ f ( x )x for
Substitute this value of m and the
all x ≠ c in the interval.
coordinates of the known point P ( x 0 , y 0 )
into the point-slope form denoted by
y− y 0=m(x−x 0)
TANGENT LINE
Example:
The tangent lines at the “peaks” and
“troughs” of a smooth curve are horizontal. Let us find the equation of the tangent line of
The tangent line drawn on ( 0 , ±1 ) and y=x 2 at x=2.
( ± 1,0 ) is shown below
Find the value of ( x 0 , y 0 ).
x 0=2
To find the value of y 0 by substituting
x 0=2 into y=x 2.
y 0=2 2=4
At ( ± 1,0 ), the tangent lines are vertical and Get the slope of the tangent line.
at ( 0 , ±1 ) , the tangent lines are horizontal. x 2−4
m=lim =4
x →2 x−2
The tangent line drawn on the points at Substitute this value of m and the
the first, second, third and fourth quadrant coordinates of the known point P ( x 0 , y 0 )
is shown below into the point-slope form.
y−4=4 ( x −2 )
y=4 x−4
DERIVATIVE
Let the graph below be the graph of a
function f (x)with P( x , f (x )) as the
coordinate. Let us locate another point on
At points in the first and third quadrants,
the graph which will be labeled as point Q
the tangent lines are slanting to the left
and at points in the second and fourth with coordinate ( x +∆ x , f ( x +∆ x )) and
quadrants, the tangent lines are slanting to connect the two lines with a secant line
the right. PQ .
1 −1 1 2√x
f ' (x )= x 2 = =
The slope of the secant line is:
2 2 √x 2 x
Q y −P y
m= Constant Multiple Rule
Qx −P x If c is a constant f ( x ) is a differentiable
f ( x +∆ x )−f ( x) dy
m= function then f ' ( x )=c f (x).
( x+ ∆ x)−x dx
f ( x +∆ x )−f ( x ) 13
m= f (x)= √x
∆x 3
1
1 ∗1 1 −1 −2
' 1 3 1 3 3 1 1
f (x )= √ x= x = x3 = x 3 = 3 2
3 3 3 9 9 √x
Sum and Difference Rule
If f ( x )=g ( x ) +h ( x )are differentiable
functions, then f ' x=g ' ( x ) ±h ' ( x ) ±…..
4
1 1 3 3
f ( x )= x + x −6 x+
5 3 4
4−1
1 1 3−1 1−1
∆ x and ∆ y will be smaller as Q f ' ( x )= (4 ) x + (3) x −6 ( 1 ) x +0
approaches P. 5 3
Since we are talking about the slope of the
4 3 2
tangent line which is the limit of the slopes ¿ x + x −6
of the secant lines PQ , we need to find the 5
limit as ∆ x approaches 0.
f ( x +∆ x )−f ( x ) Product Rule
mTL= lim
∆ x→ 0 ∆x
If f ( x )=¿ g( x )* h( x ) are differentiable
This limit is so important that it is given a functions, then f ' (x)=g ( x ) h ' ( x ) + h ( x ) g ' ( x )
name. It is called the derivative. In other f ( x )=( 3 x 2−5 x +1 ) ( 3 x2 −4 )
words, the general equation for the slope dy dy
of the tangent line is the derivative of the f ' ( x )=( 3 x 2−5 x+ 1 ) ( 3 x 2−4 ) + ( 3 x 2−4 ) ( 3 x 2−
function.
dx dx
Differentiation Rules ¿ ( 3 x 2−5 x+1 ) ( 6 x ) + ( 3 x 2−4 ) ( 3 x −5 )
Constant Rule ¿ 18 x3 −30 x2 +6 x +9 x 3−15 x 2−12 x +20
If c is any constant denoted by f ( x )=c, then ¿ 27 x 3−45 x 2−6 x +20
f ' ( x )=0
lim [ f ( x+ ∆ x ) ] −f ( x )
mTL= ∆x →0
Quotient Rule
∆x
lim 9−9 g( x )
¿ ∆x → 0 If f ( x )= are differentiable functions and
∆x h( x)
m TL=0 g ( x ) ≠ 0 , thenf ' ( x)=h(x) ¿ ¿.
( 5 x 2−3 x+ 2 ) dy (3 x2 + 4)−( 3 x 2 +4 ) dy ( 5 x 2−3 x
Power Rule
If f ( x )=x n is a power function, then dx dx
' n−1 f ' ( x )=
f (x )=n x 2
( 5 x −3 x +2 )
2
1
f (x)= x 2
= c. the graph of f (x) has a corner or
2 2
(5 x −3 x+ 2)(6 x )−( 3 x + 4 ) ( 10 x−3 ) cusp at x=a .
2
( 5 x2 −3 x +2 )
=
3 2 3 2
30 x +18 x −12 x−30 x +9 x −40 x +12
2
( 5 x 2−3 x+ 2 )
27 x2 −52 x +12
= 2
( 5 x 2−3 x+2 )
Chain Rule
'
It is in the form f ( g ( x ) )=f ( g ( x ) )∗g (x)
1
2
f ( x )=( 15 x −6 x +2 )
2
1
1 −1
dy
f ' (x)= ( 15 x 2−6 x +2 ) 2 (15 x2 −6 x+2)
2 dx
−1
1
¿ ( 15 x −6 x+2 ) 2 (30 x −6)
2
2
−1
30 x−6
¿ ( 2 )
( 15 x −6 x +2 ) 2
2
15 x−3
¿
√15 x 2−6 x +2
f ( x )=(3 x−2) ( 5 x+ 4 )2
f '¿
¿ ( 3 x−2 ) ( 2 )( 5 x+ 4 )2−1 (5)+ ( 5 x+ 4 )2(3)
¿ 10(3 x−2)(5 x+ 4)+3 ( 5 x+ 4 )2
¿ 10(15 x 2+ 2 x−8)+3( 25 x 2 +40 x +16)
¿ 150 x2 +20 x−80+ 75 x 2 +120 x+ 48
¿ 225 x 2+140 x−32
DIFFERENTIABILITY AND CONTINUITY
If f (x) is continuous at x=a , it does not
mean that ( x) is differentiable at x=a .
If f (x) is not continuous at x=a , then f is
not differentiable x=a .
If f (x) is not differentiable at x=a , it
does not mean that f (x) is not
continuous at x=a .
A function f (x) is not differentiable at x=a
if one of the following is true:
a. f (x) is not continuous at x=a .
b. the graph of f ( x ) has a vertical
tangent line at x=a
