..Two special
limits ..The
derivative of
..Formulas
to calculate the derivative of
..Partial
derivatives
..Extension
of the chain rule
..Implicit
Differentiation
..Maximum
and minimum values
..Concavity
..Theorem of
Cauchy..L'Hospitals rule
and special limits
Two special limits
Without proof, we accept that
sin(x)
lim ------- = 1
x->0 x
|
tan(x) sin(x) sin(x) 1
lim ------- = lim --------- = lim ------ .------- = 1.1 = 1
0 x 0 x.cos(x) x cos(x)
tan(x)
lim ------- = 1
x->0 x
|
Example :
lim ( cot(x) - 1/x) = lim ( 1/tan(x) - 1/x)
0 0
1 - tan(x)/x 1 - 1
= lim (--------------) = ------- = 0
0 tan(x)/x 1
The derivative of
f(x) for x = a
Say G is the graph of f(x) and point P(a,f(a)) is on G. Now, take a point Q(a+h,f(a+h)) on G close to point P.
(f(a+h)-f(a))
The slope of PQ is ---------------
h
This slope can be viewed as the mean slope of G in the curve segment
[PQ].
Now, let h become smaller and smaller. Then the line PQ approaches to
the tangent line to the curve at point P.
If the following limit exists and
if it is real, then
(f(a+h)-f(a))
the slope of that tangent line = lim --------------
h->0 h
The last limit is called the derivative of f(x) at point P. This value is
noted f'(a).
f(x)
Now make the point P from previous section variable. Then
the slope of the tangent line in point P(x,f(x))
(f(x+h)-f(x))
= lim ---------------
h->0 h
= the derivative of f(x)
= f'(x) (notation)
df
= --- (notation)
dx
d
= --- f(x) (notation)
dx
If, for x = b, this limit is not a real number, then we say that the
derivative does not exist for x = b, or that f(x) is not differentiable for x =
b... A function is differentiable in [a,b] if it is differentiable for each x in [a,b].
Differentiable and continuous
Theorem:If a function f(x) is differentiable for x = b, then f(x) is continuous for x = b.
Proof:
Since f(x) is differentiable for x = b, we have
(f(b+h)-f(b))
lim --------------- = a real number = f'(b)
h->0 h
Now,
f(b+h) - f(b)
lim f(b+h) = lim ( ---------------- .h + f(b) )
h->0 h->0 h
f(b+h) - f(b)
lim ( ---------------- .h ) + f(b)
= h->0 h
= f'(b) .0 + f(b)
= f(b)
Let x = b+h ; if h ->0 then x -> b
lim f(x) = f(b)
x->b
And f(x) is continuous in b.
Formulas to calculate the derivative of
A constant function
Take f(x) = c.The slope of the tangent line in point p of a constant function is 0. So,
d
-- c = 0
dx
|
f(x) = x
The slope of the tangent line in point p of the function x is 1. So,
d
-- x = 1
dx
|
The sum f(x) + g(x)
If the functions f(x) and g(x) are differentiable then
d f(x+h)+g(x+h) -(f(x)+g(x))
-- (f(x)+g(x)) = lim -----------------------------
dx h->0 h
(f(x+h)-f(x)) + (g(x+h)-g(x))
= lim -----------------------------
h->0 h
(f(x+h)-f(x)) (g(x+h)-g(x))
= lim --------------- + lim ---------------
h->0 h h->0 h
d d
= -- f(x) + -- g(x)
dx dx
d d d -- (f(x)+g(x)) = -- f(x) + -- g(x) = f'(x) + g'(x) dx dx dx |
This property is extendable to the sum of n differentiable functions.
The product f(x).g(x)
If the functions f(x) and g(x) are differentiable then
d f(x+h).g(x+h) - f(x).g(x)
-- (f(x).g(x)) = lim ---------------------------
dx h->0 h
f(x+h).g(x+h) -f(x+h)g(x) + f(x+h)g(x) - f(x).g(x)
= lim --------------------------------------------------
h->0 h
(g(x+h)-g(x)) (f(x+h)-f(x))
= lim (f(x+h). --------------- + g(x). --------------)
h->0 h h
= f(x).g'(x) + f'(x).g(x)
d -- (f(x).g(x)) = f(x).g'(x) + f'(x).g(x) dx |
The product of n differentiable functions.
Previous property is extendable to the product of n differentiable functions.Example : u=f(x) v=g(x) w=h(x)
d --(u.v.w) = u'.v.w + u.v'.w + u.v.w' dx |
Special product
u=f(x) and c is a constant
d --(c.u) = c.u' dx |
Power of f(x)
u=f(x) is differentiabled n d n-1 --( u ) = --(u.u.u. ... .u) = n.(u ).u' dx dx
d --( un ) = n.(un-1).u' dx |
The quotient f(x)/g(x)
If the functions f(x) and g(x) are differentiable then
d f(x+h)/g(x+h) -(f(x)/g(x))
-- (f(x)/g(x)) = lim -----------------------------
dx h->0 h
f(x+h).g(x) - f(x).g(x+h)
= lim -----------------------------
h->0 h.g(x).g(x+h)
f(x+h).g(x) -f(x)g(x) +f(x)g(x) - f(x).g(x+h)
= lim ---------------------------------------------
h->0 h.g(x).g(x+h)
g(x) . (f(x+h)-f(x))/h - f(x) . (g(x+h)-g(x))/h
= lim ------------------------------------------------
h->0 g(x).g(x+h)
g(x).f'(x) - f(x).g'(x)
= -------------------------
g(x).g(x)
(u / v)' = ( v u' - u v') / v2
|
Important special case
u=f(x) is differentiable
n (n-1)
d -n d 1 u . 0 - n.u .u' (-n-1)
--( u ) = -- (---) = --------------------- = -n.u .u'
dx dx n 2n
u u
From this it follows that the formula d n n-1 --( u ) = n.(u ).u' dxholds for all integer values of n.
sin(x)
d sin(x+h) - sin(x)
--sin(x) = lim ------------------
dx h->0 h
2.cos(x + h/2).sin(h/2)
= lim ------------------------
h->0 h
cos(x + h/2).sin(h/2)
= lim ------------------------
h->0 h/2
sin(h/2)
= lim cos(x + h/2).--------
h->0 h/2
= cos(x)
cos(x)
Analogous as for sin(x) you can prove thatd --cos(x) = - sin(x) dx
tan(x)
d d sin(x) cos2(x) - sin(x).(-sin(x))
--tan(x) = --(------) = -----------------------------
dx dx cos(x) cos2(x)
1
= ---------
cos2(x)
cot(x)
Analogous as for tan(x) you can prove thatd -1 --cot(x) = --------- dx sin2(x)
The chain rule
Take the function f(g(x)) en let u = g(x). Assume that f and g are differentiable.
d f(g(x+h)) - f(g(x))
--(f(g(x)) = lim ------------------
dx h->0 h
f(g(x+h)) - f(g(x)) g(x+h) - g(x)
= lim --------------------. ----------------
h->0 g(x+h) - g(x) h
let g(x+h) = g(x) + k = u + k
f(u + k) - f(u) g(x+h) - g(x)
= lim -------------------. ----------------
h->0 k h
if h -> 0 then k -> 0
f(u + k) - f(u) g(x+h) - g(x)
= lim -------------------. lim ----------------
k->0 k h->0 h
d d
= -- f(u) . -- g(x)
du dx
We can also write
d d -- f(u) = -- f(u) . u' dx duThis formula is known as the chain rule. |
Corollaries
Appealing on the chain rule we haved d --sin(u) = -- sin(u) . u' = cos(u) . u' dx du d d --cos(u) = -- cos(u) . u' = - sin(u) . u' dx du d d 1 --tan(u) = -- tan(u) . u' = ---------- . u' dx du cos2(u) d d 1 --cot(u) = -- cot(u) . u' = - -------- . u' dx du sin2(u) d --(un ) = n . (un-1).u' dx ...
d --sin(u) = cos(u) . u' dx d --cos(u) = - sin(u) . u' dx d 1 --tan(u) = ---------- . u' dx cos2(u) d 1 --cot(u) = - -------- . u' dx s in2(u) d --(un ) = n . (un-1).u' dx |
The inverse trigonometric functions
First take arcsin(x) . We know that for x in [-1,1] sin( arcsin(x) ) = x. (1)So, these two functions must have the same derivative.
Thus, we differentiate both sides of (1).
From previous formulas we have
d
cos( arcsin(x) ). --( arcsin(x) ) = 1 (2)
dx
But let b = arcsin(x). Then b is in [-pi/2,+pi/2] .
_____________ _______
| 2 | 2
cos(b) = \| 1 - sin (b) = \| 1 - x
________
| 2
Thus, cos( arcsin(x) ) = \| 1 - x
From (2) we can write now :
________
| 2 d
\| 1 - x . --( arcsin(x) ) = 1
dx
d 1
<=> --( arcsin(x) ) = -------------
dx _______
| 2
\| 1 - x
and with the chain rule :
d 1
--( arcsin(u) ) = -------------.u'
dx _______
| 2
\| 1 - u
|
Analogous
d - 1
--( arccos(u) ) = -------------.u'
dx _______
| 2
\| 1 - u
d 1
--( arctan(u) ) = -------------.u'
dx 1 + x2
d - 1
--( arccot(u) ) = -------------.u'
dx 1 + x2
|
Rational power of u = f(x)
The n-th root of x
For x > 0 we have
(x(1/n))n= x
So, these two functions must have the same derivative.Thus, we differentiate both sides.
(1/n) n-1 d (1/n)
n.(x ) . -- (x ) = 1
dx
<=> ...
d (1/n) 1 1/n - 1
<=> -- (x ) = ---. (x )
dx n
The n-th root of u = f(x)
From previous formula we have
d (1/n) 1 1/n - 1
-- (u ) = ---. (u ).u'
dx n
Rational power of x
d (m/n) d (1/n) m (1/n) m-1 1 1/n - 1
-- (x ) = -- (x ) = m.(x ) . --. (x )
dx dx n
<=> ...
m
<=> = --- . xm/n - 1
n
A rational power of u = f(x)
From previous formula we have
d (m/n) m
-- (u ) = --- . um/n - 1.u'
dx n
Logarithmic functions
Exponential functions
Real power of x
uv
Partial derivatives
definition of partial derivatives
Suppose we have a function f such that the image depends on several independent variables, say x, y and z.We write f(x,y,z).
Consider, for a moment, in such function f(x,y,z) y and z as constant, then
f(x,y,z) only depends on the variable x. Then, we can calculate the derivative
with respect to x.
We note this derivative as
fx'(x,y,z)
Similarly we can calculate
fy'(x,y,z) and fz'(x,y,z)
Example of partial derivatives
f(x,y,z) = 3 x y - x z + 6 y -4
fx'(x,y,z) = 3 y - z
fy'(x,y,z) = 3 x + 6
fz'(x,y,z) = - x
Extension of the chain rule
Suppose we have a function f(x,y,z) such that x, y and z are not independent but functions of a variable t. Then f is a function of t.Suppose that x, y and z are differentiable functions of t. We denote the three derivatives for short as x',y' and z'. We can calculate the derivative of f with respect to t with the formula
d
--- f(x,y,z) = fx'(x,y,z) . x' + fy'(x,y,z) . y' + fz'(x,y,z) . z'
dt
|
The proof of this formula is beyond the scope of this tutorial.
Example:
f(x,y,z) = x2 + x.y + z2 + y.z
and x = 3t ; y = 5 t2 ; z = t3
Then
d
--- f(x,y,z) = (2x + y).3 + (x + z). 10t + (2z + y). 3t2
dt
= (6t + 5t2).3 + (3t + t3) .10t + (2t3 + 5t2).3t2
= ...
= 6 t5 + 25 t4 + 45 t2 + 18 t
Implicit Differentiation
Implicit functions
Example :
We define the function
x + 6
y = -----
x - 5
If x is not 5, the same function is defined by one the implicit forms y.(x - 5) = x + 6 or x y = x + 5 y + 6 or x y - x - 5 y - 6 = 0
Implicit definition of a function
From previous example we see that expressions as
g(x,y) = 0
or g(x,y) = h(x,y)
can define y implicitly as a function of x.
Implicit differentiation
Concept of implicit differentiation
Suppose that we know that y is implicitly defined as a function of x by the expression
g(x,y) = h(x,y)
If we think y as a function of x, the previous form expresses an identity
for all x-values.Then, the left side is a function of x, and the right side is a function of x and these functions are identical.
Therefore the derivative, with respect to x, of both sides is the same.
d g(x, y) d h(x, y) --------- = --------- dx dxThis defines implicitly y'.
Example of implicit differentiation
From previous example we know that
x y = x + 5 y + 6
defines y implicitly as a function of x.If we think y as a function of x, the previous form expresses an identity for all x-values.
Then, the left side is a function of x, and the right side is a function of x and these functions are identical.
Therefore the derivative, with respect to x, of both sides is the same.
1.y + x.y' = 1 + 5 y' + 0
<=>
(x-5) y' = 1 - y
<=>
1 - y
y' = -----
x - 5
Previous procedure to calculate y' is called implicit differentiation
Generalization
Previous procedure can be generalized to calculate y' even if the expression
g(x,y) = h(x,y)
defines two or more different y values as a function of x.Example:
2 y = 4x defines two different y values as a function of x.Thinking of y as a function of x, we obtain
2 y y' = 4
<=>
y' = 2/y
This result holds for the two different functions!!
Maximum and minimum values
Relative maximum
We say that a function f(x) has a relative maximum for x = t if and only if there is a strictly positive real number e such that f(x) =< f(t) for all x in ]t-e,t+e[ .Often the word 'relative' is omitted.
Relative minimum
We say that a function f(x) has a relative minimum for x = t if and only if there is a strictly positive real number e such that f(x) >= f(t) for all x in ]t-e,t+e[ .Often the word 'relative' is omitted.
Roots of f'(x)
| If f'(x) exists in an open interval that contains the value x = t, and
suppose that f reaches a maximum for x = t.
Then, f'(t) = 0. |
Proof:
f'(x) exists for x = t
(f(t+h)-f(t))
=> lim --------------- = a number g
h->0 h
We consider the right and left limit separately
(f(t+h)-f(t)) (f(t+h)-f(t))
=> lim --------------- = g and lim --------------- = g
> 0 h < 0 h
Since f(x) reaches a maximum for x = t , we have
=> ( g > or = 0 ) and ( g < or = 0 )
=> g = 0
=> f'(t) = 0
In the same way we can prove :
| If f'(x) exists in an open interval that contains the value x = t, and
suppose that f reaches a minimum for x = t.
Then, f'(t) = 0. |
Rolle's theorem
If
|
Proof:
If f is constant in [a,b], then the proof is trivial.
Now, suppose f is not constant in [a,b].
Then there is a number d in ]a,b[ such that f(d) is different from f(a) and f(b).
Suppose first that f(d) > f(a). Since f is continuous in [a,b], f attains, according to Weierstrass, a greatest image.
Thus, there is a c-value in ]a,b[ such that f(c) is maximum and since f is differentiable in ]a,b[, f'(c) = 0.
In the same way, you can prove the theorem for f(d) < f(a).
Lagrange's theorem
If
f(b) - f(a)
f'(c) = ---------------
b - a
|
C is the graph of y = f(x).
The slope of PQ is ( f(b) - f(a) ) / (b -
a).
f'(c) is the slope of the tangent line in point R(c,f(c)).
The theorem
says that there is a suitable point R on the curve C such that the tangent line
in R is parallel to PQ.
Proof:
The equation of PQ is
f(b) - f(a)
y - f(a) = -------------- (x - a)
b - a
<=>
f(b) - f(a)
y = f(a) + -------------- (x - a)
b - a
Now, we consider three functions
f : x --> y = f(x)
f(b) - f(a)
g : x --> y = f(a) + -------------- (x - a)
b - a
h : x --> y = f(x) - g(x)
It is not difficult to verify that the h(x) satisfies the three conditions
of Rolle's theorem.
We apply Rolle's theorem on the function h(x). There is a c-value in ]a,b[ such that h'(c) = 0;
h'(x) = f'(x) - g'(x)
f(b) - f(a)
= f'(x) - 0 - ------------- .1
b - a
There is a c-value in ]a,b[ such that
f(b) - f(a)
f'(c) - ------------- = 0
b - a
<=>
f(b) - f(a)
f'(c) = ------------
b - a
Constant functions
Theorem:If f'(x) = 0 for all x in [a,b]
Then f(x) is constant in [a,b].
Proof:
Take two values c and d in [a,b]. We'll show that f(c) = f(d).
Since f'(x) exists for each x in [a,b], f'(x) exists in [c,d] and f(x) is
continuous in [c,d].
We apply lagrange's-theorem in [c,d].
There is a
value e in ]c,d[ such that
f(c) - f(d)
f'(e) = -------------
c - d
but f'(x) = 0 for all x in [c,d]. Thus,
f(c) - f(d)
------------- = 0
c - d
and f(c) = f(d)
Increasing functions
Theorem:If f'(x) > 0 for all x in [a,b]
Then f(x) is increasing in [a,b].
Proof:
Take two values c < d in [a,b]. We'll show that f(c) < f(d).
Since f'(x) exists for each x in [a,b], f'(x) exists in [c,d] and f(x) is
continuous in [c,d].
We apply lagrange's-theorem in [c,d].
There is a
value e in ]c,d[ such that
f(c) - f(d)
f'(e) = -------------
c - d
but f'(x) > 0 for all x in [c,d]. Thus,
f(c) - f(d)
------------- > 0
c - d
Since c < d , we have f(c) < f(d).
Decreasing functions
Theorem:If f'(x) < 0 for all x in [a,b]
Then f(x) is decreasing in [a,b].
(proof similar to previous theorem)
Relative maximum of a differentiable functions.
Let e = a strictly positive real number.Suppose that a function is differentiable in ]t-e,t+e[ .
If f'(x) > 0 in a suitable interval ]t-e,t[ , the f(x) is increasing at
the left side of t.
If f'(x) < 0 in a suitable interval ]t,t+e[ , the f(x)
is decreasing at the right side of t.
In that case f(x) has a relative
maximum for x = t.
Relative minimum of a differentiable functions.
Let e = a strictly positive real number.Suppose that a function is differentiable in ]t-e,t+e[ .
If f'(x) < 0 in a suitable interval ]t-e,t[ , the f(x) is decreasing at the left side of t.
If f'(x) > 0 in a suitable interval ]t,t+e[ , the f(x) is increasing at the right side of t.
In that case f(x) has a relative minimum for x = t.
Detection of a relative maximum or relative minimum
In general we detect a relative maximum or relative minimum of f(x) by investigating the sign of f'(x).| f(x) has a maximum = f(t) if f'(x) changes from + to - for x=t.
f(x) has a minimum = f(t) if f'(x) changes from - to + for x=t. |
Example
f(x) = 3x5- 5x3 then f'(x) = 15x4-15x2= 15 x2.(x-1)(x+1)
Investigation of the sign gives
x | -1 0 1
---------------------------------------
f'(x) | + - - +
So there is a maximum for x = -1 and a minimum for x = 1
Concavity
Concave upward
If, in an interval, f"(x) > 0 , then f'(x) is increasing.Then the slope of the tangent line is increasing with x.
We say that f(x) is concave upward in that interval.
| If, in an interval, f"(x) > 0 then f(x) is concave upward in that interval. |
Concave downward
If, in an interval, f"(x) < 0 , then f'(x) is decreasing. Then the slope of the tangent line is decreasing with x. We say that f(x) is concave downward in that interval.
| If, in an interval, f"(x) < 0 then f(x) is concave downward in that interval. |
Points of inflection
Example:
f(x) = 3x5- 5x3 then f"(x) = 15(x4 - x2) = 30x(2x2- 1)
Investigation of the sign gives
x | -sqrt(1/2) 0 sqrt(1/2)
----------------------------------------------------
f"(x) | - + - +
So,
for x < -sqrt(1/2) the graph is concave downward.
for -sqrt(1/2) < x < 0 the graph is concave upward.
for 0 < x < sqrt(1/2) the graph is concave downward.
for x > sqrt(1/2) the graph is concave upward.
The points where the concavity changes sign are the points with
x = -sqrt(1/2) ; x = 0 ; x = sqrt(1/2).
These points are called points of inflection.
| The points where the concavity changes sign are the points of inflection. |
Theorem of Cauchy
If f(x) and g(x) are continious in [a,b] and if f'(x) and g'(x) exist
in ]a,b[ and have no common roots in ]a,b[ , then there is a value c in
]a,b[ such that
f(b) - f(a) f'(c)
------------- = -------
g(b) - g(a) g'(c)
|
Proof:
Denote
f(b) - f(a)
------------- = K (1)
g(b) - g(a)
Then f(b) - f(a) - K (g(b) - g(a)) = 0
We create the function f(x) - K g(x).This function is continious in [a,b] and the derivative exists in ]a,b[. On this function, we can use Lagrange's theorem .
There is a value c in ]a,b[ such that
f(b) - K g(b) - ( f(a) - K g(a) ) = (b-a).(f'(c) - K g'(c))
=> 0 = (f'(c) - K g'(c))
If g'(c) = 0 then f'(c) = 0 and this is impossible because f'(x) and g'(x)
have no common roots in ]a,b[. So, g'(c) is not 0 and then
f'(c)
------ = K (2)
g'(c)
From (1) and (2), we have that there is a value c in ]a,b[ such that
f(b) - f(a) f'(c)
------------- = -------
g(b) - g(a) g'(c)
L'Hospitals rule and special limits
Theorem 1
If ( lim f(x) = lim g(x) = 0 )
a a
f'(x)
and lim ------- = can be found
a g'(x)
Then
f(x) f'(x)
lim ------ = lim -------
a g(x) a g'(x)
|
Proof :
f'(x)
First, suppose that lim ------- = finite = A
a g'(x)
Since f'(x) exists in the environment of a, f(x) is continuous in the
environment of a and
lim f(x) = f(a)
a
But we have also that
lim f(x) = 0
a
Therefore f(a) = 0.
Similarly g(a) = 0.With Cauchy's theorem, we can write in the same environment of a
f(x) f(x) - f(a) f'(c)
----- = ------------- = ------- with c between x and a.
g(x) g(x) - g(a) g'(c)
If x --> a , c --> a.Now we take the limit of both sides for x --> a
f(x) f'(c)
lim ------ = lim ------- = A
a g(x) a g'(c)
And, in this first case, the theorem is proved.
f'(x)
Now, suppose that lim ------- = + infinity
a g'(x)
Then
g'(x)
lim ------- = +0
a f'(x)
and appealing on the first case
g(x) g'(c)
lim ------ = lim ------- = +0
a f(x) a f'(c)
So,
f(x) f'(c)
lim ------ = lim ------- = + infinity
a g(x) a g'(c)
(similar for - infinity)
Theorem 2
If ( lim f(x) = lim g(x) = infinite )
a a
f'(x)
and lim ------- = can be found
a g'(x)
Then
f(x) f'(x)
lim ------ = lim -------
a g(x) a g'(x)
|
Proof:
f'(x)
First, suppose that lim ------- = finite = A
a g'(x)
Consider the identity for all x
f(x) f(x) - f(b) 1 - g(b)/g(x)
----- = -------------. ---------------
g(x) g(x) - g(b) 1 - f(b)/f(x)
Since f'(x) exists in the environment of a, f(x) is continuous in the
environment of a and similarly for g(x).
With Cauchy's theorem, previous identity becomes, with b and x in the same environment of a :
f(x) f'(c) 1 - g(b)/g(x)
----- = ------- . --------------- with c between x and a.
g(x) g'(c) 1 - f(b)/f(x)
If x --> a , c --> a.Now we take the limit of both sides for x --> a
f(x) f'(c) 1 - 0
lim ------ = lim -------. ------ = A
a g(x) a g'(c) 1 - 0
And, in this first case, the theorem is proved.
f'(x)
Now, suppose that lim ------- = + infinity
a g'(x)
Then
g'(x)
lim ------- = +0
a f'(x)
and appealing on the first case
g(x) g'(c)
lim ------ = lim ------- = +0
a f(x) a f'(c)
So,
f(x) f'(c)
lim ------ = lim ------- = + infinity
a g(x) a g'(c)
(similar for - infinity)
Special limits - examples
L'Hospitals rule is frequently used in the following limits.In these examples the properties and the derivatives of exponential and logarithmic functions are used.
cos(2x) - 1 -2 sin(2x) lim ------------ = lim ------------ = 0 0 sin(3x) 0 3 cos(3x)cos(x)-cos(2x) - sin(x) + 2 sin(2x) lim -------------- = lim --------------------- 0 cos(x)-cos(3x) 0 - sin(x) + 3 sin(3x) - cos(x) + 4cos(2x) 3 = lim -------------------- = --- 0 - cos(x) + 9cos(3x) 8ln(x) 1/x lim ------- = lim ------- = 0 infty x infty 1xn n.xn-1 lim ------ = lim --------- infty ex ex n.(n-1)xn-2 = lim -------------- ex = ... n! = lim ------ = 0 exln(x) 1/x lim x. ln(x) = lim ------ = lim ------ = - lim x = 0 0 1/x -1/x2lim xx = lim ex.ln(x) = elim x.ln(x) 0 0 and from previous example = e0 = 1lim (cos(x))1/x 0 = lim e(1/x).ln(cos(x)) 0 = elim (1/x).ln(cos(x)) but lim (1/x).ln(cos(x)) 0 ln(cos(x)) = lim --------------- 0 x -sin(x)/cos(x) = lim ---------------- = 0 0 1 So lim (cos(x))1/x = e0 = 1 0
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