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 
sin(5x) lim  = 1 0 5x tan(53x) lim  = 1 0 53x 5x lim  = 1 0 sin(5x) 3x (3/5) (5x) lim  = lim  = 3/5 0 sin(5x) 0 sin(5x) 5 x^{2} (1/5) (5x) (5x) lim  = lim  = 1/5 0 sin^{2}(5x) 0 sin(5x) sin(5x) sin(5x) sin(5x) / (5x) lim  = lim  = 1 0 tan(5x) 0 tan(5x) / (5x) sin(3x) 3 sin(3x) / (3x) lim  = lim  = 3/2 0 tan(2x) 0 2 tan(2x) / (2x) tan^{2}(4x) 16 (tan(4x)/(4x)) (tan(4x)/(4x)) lim  = lim  = 16 0 x sin(x) sin(x)/x 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 1All these limits can also be calculated using de l'Hospital rule. See below.
(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 hThe last limit is called the derivative of f(x) at point P. This value is noted f'(a).
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) = D(f(x)) (notation) df =  (notation) dx d =  f(x) (notation) dxIf, for x = b, this limit does not exist or if it 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].
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.
d  c = 0 or D(c) = 0 dx 
Examples: D 3 = 0 ; D sin(4) = 0
d  x = 1 or D(x) = 1 dx 
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 
Example: D(x + 1/pi) = 1 + 0 = 1
This property is extendable to the sum of n differentiable functions.
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 
d (u.v.w) = u'.v.w + u.v'.w + u.v.w' dx 
d (c.u) = c.u' dx 
Examples:
D(3x) = 3 D(2x + 13) = 2 D(x^{2}/434) = 1/434 D(x^{2})
d n d n1 ( u ) = (u.u.u. ... .u) = n.(u ).u' dx dx
d ( u^{n} ) = n.(u^{n1}).u' dx 
Examples:
D(x^{3}) = 3 x^{2} D(x^{8}) = 8 x^{7} D(x^{3} + x^{4}) = 3x^{2} + 4x^{3} D(x^{3} + 34) = 3 x^{2} D( (12x)^{3} ) = 3(12x)^{2}.(12) D( (12x + 4)^{4} ) = 4(12x +4)^{3}.12 D( (12x+4)(x^{4}+5)) = D(12x+4) .(x^{4}+5) + (12x+4).D(x^{4}+5) = 12.(x^{4}+5) + (12x+4).4x^{3} D( (x^{3}.(x+1)^{5}) = 3x^{2}.(x+1)^{5} + x^{3}.5(x+1)^{4} D( (x^{3}.(x^{2}+1)^{5}) = 3x^{2}.(x^{2}+1)^{5} + x^{3}.5(x^{2}+1)^{4}.2x
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') / v^{2} 
Example :
d 2x+7 (3x+2)2  (2x+7)3  () =  dx 3x+2 (3x+2)^{2}
n (n1) d n d 1 u . 0  n.u .u' (n1) ( u ) =  () =  = n.u .u' dx dx n 2n u uFrom this it follows that the formula
d n n1 ( u ) = n.(u ).u' dxholds for all integer values of n.
Here are some additional useful formulas:
D( 1/x) = D( x^{1} ) = 1.x^{2} = 1/x^{2} d d (1/u ) = (u^{1} ) = 1.u^{2}.u' dx dx 1 =  u' u^{2}
d 1 (1/u ) =  u' dx u^{2} 
Examples:
D( 1/(2x+23) ) = 2/(x+23)^{2} 1 1 D( ) =  .4. 5x^{4} 4x^{5} (4x^{5})^{2} 1 1 D  =  . 4(2x+23)^{3}. 2 (2x+23)^{4} (2x+23)^{8} D (x^{2}+4x)^{3} = 3(x^{2}+4x)^{2} (2x+4) D ((x1)(x+1)(x^{2}+1)) = D ((x^{2}1)(x^{2}+1)) =D (x^{4}1) = 4x^{3} D (1/(2x+4)^{3}) = D (2x+4)^{3} = 3 (2x+4)^{4}.2 = 6/(2x+4)^{4}
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)
d cos(x) =  sin(x) dx
d d sin(x) cos^{2}(x)  sin(x).(sin(x)) tan(x) = () =  dx dx cos(x) cos^{2}(x) 1 =  cos^{2}(x)
d 1 cot(x) =  dx sin^{2}(x)
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 dxWe can also write
d d  f(u) =  f(u) . u' dx duThis formula is known as the chain rule. 
d 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 cos^{2}(u) d d 1 cot(u) =  cot(u) . u' =   . u' dx du sin^{2}(u) d (u^{n} ) = n . (u^{n1}).u' dx ...
d sin(u) = cos(u) . u' dx d cos(u) =  sin(u) . u' dx d 1 tan(u) =  . u' dx cos^{2}(u) d 1 cot(u) =   . u' dx sin^{2}(u) d (u^{n} ) = n . (u^{n1}).u' dx 
Examples:
D( sin(x^{2}) ) = cos(x^{2}) . 2x 3 D( tan(3x+2) ) =  cos^{2}(3x+2) D ( x^{2} . sin(3x) ) = 2x . sin(3x) + x^{2}. cos(3x).3 D ( sin(x)/x) = D( (1/x).sin(x)) = (1/x^{2}).sin(x) +(1/x).cos(x) 1 D tan(sqrt(x^{3})) = . (3/2) x^{1/2} cos^{2}(x^{3/2}) 1 D tan(sin(1/x)) =  cos(1/x) . (1/x^{2}) cos^{2}(sin(1/x)) D sin^{3}(pi/2  x) = D cos^{3}(x) = 3cos^{2}(x) (sin(x)) 1 D  = D cos^{2}(x) = 2 cos^{3}(x).(sin(x)) cos^{2}(x) D sin^{m}(x^{m}) = m sin^{m1}(x^{m}) cos(x^{m}) m x^{m1} x^{2}  1 x^{2}  1 (x^{2} + 1)2x (x^{2}  1)2x D sin() = cos() .  = ... x^{2} + 1 x^{2} + 1 (x^{2} + 1)^{2}
d cos( arcsin(x) ). ( arcsin(x) ) = 1 (2) dxBut 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  xFrom (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 
d  1 ( arccos(u) ) = .u' dx _______  2 \ 1  u d 1 ( arctan(u) ) = .u' dx 1 + u^{2} d  1 ( arccot(u) ) = .u' dx 1 + u^{2} 
Examples:
d  1 ( arccos(x^{7}) ) =  7 x^{6} dx ________  14 \ 1  x d 1 ( arcsin(1/x)) =  (1/x^{2}) dx ___________  2 \ 1  (1/x) d 1 ( arctan(sin(2x)) ) =  (cos(2x)).2 dx 1 + sin^{2}(2x)
(x^{(1/n)})^{n}= xSo, these two functions must have the same derivative.
(1/n) n1 d (1/n) n.(x ) .  (x ) = 1 dx <=> ... d (1/n) 1 1/n  1 <=>  (x ) = . (x ) dx n
d (1/n) 1 1/n  1  (u ) = . (u ).u' dx n
d (1/2) 1 1/2  (u ) = . (u ).u' dx 2 So, 1 D( sqrt(u) ) =  u' 2.sqrt(u)
d (m/n) d (1/n) m (1/n) m1 1 1/n  1  (x ) =  (x ) = m.(x ) . . (x ) dx dx n <=> ... m <=> =  . x^{m/n  1} n
d (m/n) m  (u ) =  . u^{m/n  1}.u' dx n
d (m/n) m  (u ) =  . u^{m/n  1}.u' dx n d 1  ( sqrt(u) ) =  u' dx 2.sqrt(u) 
Examples:
6x D( sqrt(3x^{2}+5) ) =  2. sqrt(3x^{2}+5) D( (sin(3x))^{4/3} ) = (4/3)(sin(3x))^{1/3} . cos(3x) . 3 cos(x)cos(2x) 2sin(x)sin(2x) D( sqrt(sin(x).cos(2x)) ) =  2 sqrt(sin(x).cos(2x)) 2x+1 (2x+1).   sqrt(x^{2}+x+1). 2 sqrt(x^{2}+x+1) 2.sqrt(x^{2}+x+1) D  =  = .... 2x + 1 (2x+1)^{2} 2x 2x sqrt(x^{2}+4).   sqrt(x^{2}2)  sqrt(x^{2}2) 2.sqrt(x^{2}2) 2.sqrt(x^{2}+4) D  =  = ... sqrt(x^{2}+4) (x^{2}+4)
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
f_{x}'(x,y,z)Similarly we can calculate
f_{y}'(x,y,z) and f_{z}'(x,y,z)
f(x,y,z) = 3 x y  x z + 6 y 4 f_{x}'(x,y,z) = 3 y  z f_{y}'(x,y,z) = 3 x + 6 f_{z}'(x,y,z) =  x
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) = f_{x}'(x,y,z) . x' + f_{y}'(x,y,z) . y' + f_{z}'(x,y,z) . z' dt 
f(x,y,z) = x^{2} + x.y + z^{2} + y.z and x = 3t ; y = 5 t^{2} ; z = t^{3} Then d  f(x,y,z) = (2x + y).3 + (x + z). 10t + (2z + y). 3t^{2} dt = (6t + 5t^{2}).3 + (3t + t^{3}) .10t + (2t^{3} + 5t^{2}).3t^{2} = ... = 6 t^{5} + 25 t^{4} + 45 t^{2} + 18 t
x + 6 y =  x  5If 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
g(x,y) = 0 or g(x,y) = h(x,y)can define y implicitly as a function of x.
g(x,y) = h(x,y)If we think y as a function of x, the previous form expresses an identity for all xvalues.
d g(x, y) d h(x, y)  =  dx dxThis defines implicitly y'.
x y = x + 5 y + 6defines y implicitly as a function of x.
1.y + x.y' = 1 + 5 y' + 0 <=> (x5) y' = 1  y <=> 1  y y' =  x  5Previous procedure to calculate y' is called implicit differentiation
g(x,y) = h(x,y)defines two or more different y values as a function of x.
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/yThis result holds for the two different functions!!
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. 
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) = 0In 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. 
The function f(x) is given
If

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 maximum image.
So, there is a cvalue 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).
As soon as at least one of the three conditions is not satisfied, the existence of the cvalue is not guaranteed.
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  aNow, 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 cvalue in ]a,b[ such that h'(c) = 0;
h'(x) = f'(x)  g'(x) f(b)  f(a) = f'(x)  0   .1 b  aThere is a cvalue in ]a,b[ such that
f(b)  f(a) f'(c)   = 0 b  a <=> f(b)  f(a) f'(c) =  b  a
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'stheorem 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]. So, f(c)  f(d)  = 0 c  d and f(c) = f(d)
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'stheorem 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]. So, f(c)  f(d)  > 0 c  d Since c < d , we have f(c) < f(d).
(proof similar to previous theorem)
If f'(x) > 0 in a suitable interval ]te,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.
If f'(x) < 0 in a suitable interval ]te,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.
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. 
f(x) = 3x^{5} 5x^{3} then f'(x) = 15x^{4}15x^{2}= 15 x^{2}.(x1)(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
As 1 revolution takes just one second, u = 2.pi.t
ds 2 pi 6pi v =  = 3  =  dt cos^{2}(2 pi t) cos^{2}(2 pi t) If u = 2 pi t = pi/3 then v = 24 pi m/s
Area of the sphere = A = 4 pi r^{2}
Volume of the sphere = V = (4/3) pi r^{3}
dA/dt = 8 pi r dr/dt (*) DV/dt = 4 pi r^{2} dr/dt We know that DV/dt = 1 So, 4 pi r^{2} dr/dt = 1 => dr/dt = 1/(4 pi r^{2}) (**) From (*) and (**) it follows that dA/dt = 8 pi r . ( 1/(4 pi r^{2}) ) = 2/r On the moment that r = 20, dA/dt = 0.1 The area decreases with 0.1 cm^{2}/s when the radius r is 20 cm
Let h = the water level at a given time t. Let r = the radius of water surface at a given time t. V = Volume water at a given time t. V = pi r^{2} h / 3. r and h are functions of t. d V/dt = (pi/3) ( 2 r (dr/dt) h + (dh/dt) r^{2} ) (*) But we know that dV/dt = 0.1 The figure it follows that: r/h = 2/1 ; so, r = 2h and dr/dt =2 dh/dt (*) becomes: 0.1 = (pi/3) (12 h^{2} )(dh/dt) At the moment that h = 0.5 we have dh/dt = 0.03 At the time that the water level is 0.5 dm, the water level falls 3 mm/s.
We choose to axes as shown in the figure.
Let P_{1}(0,a) P_{2}(c,b) S(x,0)
The total time T to go from A to B is a function of x.
T = P_{1} S/v_{1} + S P_{2}/v_{2} sqrt(a^{2}+x^{2}) sqrt( (cx)^{2} + b^{2} ) =  +  v_{1} v_{2} The total time T should be minimum. dT 2 x 2 (cx)(1)  =  +  dx v_{1} 2 sqrt(a^{2}+x^{2}) v_{2} 2 sqrt((cx)^{2} + b^{2} ) T is minimum if and only if dT/dx = 0. Then we have x (cx)  =  (*) v_{1} sqrt(a^{2}+x^{2}) v_{2} sqrt((cx)^{2} + b^{2} ) The angles i and j are marked on the figure. In the right triangles on the figure we have x  = sin(i) sqrt(a^{2}+x^{2}) and (cx)  = sin(j) sqrt((cx)^{2} + b^{2} ) So, the condition (*) becomes sin(i) sin(j)  =  v_{1} v_{2} sin(i) v_{1}  =  sin(j) v_{2} And we find Snell's law !
If, in an interval, f"(x) > 0 then f(x) is concave upward in that interval. 
If, in an interval, f"(x) < 0 then f(x) is concave downward in that interval. 
f(x) = 3x^{5} 5x^{3} then f"(x) = 15(x^{4}  x^{2}) = 30x(2x^{2} 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. 
If f(x) and g(x) are continuous 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) 
Denote f(b)  f(a)  = K (1) g(b)  g(a) Then f(b)  f(a)  K (g(b)  g(a)) = 0We create the function f(x)  K g(x).
There is a value c in ]a,b[ such that
f(b)  K g(b)  ( f(a)  K g(a) ) = (ba).(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)
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) 
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.
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.
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)
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) 
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.
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)
sqrt(x3) 1 lim  4 x  4 l'Hospital 1 = lim  4 2 sqrt(x3) = 1/2
______________  2 1  \ x  3 x + 3 lim  1 _________  2 \ 4 x  3  1 l'Hospital _________  2 2 \ 4 x  3 . (2x3) = lim  1 _____________  2 2 \ x  3 x + 3 . (8x) = 1/8
________  2 \ x + 1 + 3 x lim  +infty 2x  5 l'Hospital x  + 3 ________  2 \ x + 1 = lim  +infty 2 x = (1/2) lim (  ) + 3/2 +infty ________  2 \ x + 1   x^{2} = (1/2) lim  + 3/2 = 2 +infty  2 \ x + 1
lim ( sqrt(4x^{2} 4x) 2x ) +infty = 2 lim ( sqrt(x^{2}  x)  x) +infty = 2 lim x ( sqrt(1  1/x) 1) +infty sqrt(1  1/x) 1 = 2 lim  (case 0/0) +infty 1/x (1/x^{2}) = 2 lim  +infty 2 sqrt(1  1/x) . (1/x^{2}) 1 = lim  = 1 +infty sqrt(1  1/x)
lim ( (8x^{3}+1)^{1/3}  2x ) +infty = lim x ( (8 + 1/x^{3})^{1/3}  2 ) +infty (8 + 1/x^{3})^{1/3}  2 = lim  (case 0/0) +infty 1/x (3/x^{4}) = lim  +infty 3 (8 + 1/x^{3})^{2/3} (1/x^{2}) 1 = lim  = 0 +infty (8 + 1/x^{3})^{2/3} x^{2}
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) 8
ln(x) 1/x lim  = lim  = 0 infty x infty 1
x^{n} n.x^{n1} lim  = lim  infty e^{x} e^{x} n.(n1)x^{n2} = lim  e^{x} = ... n! = lim  = 0 e^{x}
ln(x) 1/x lim x. ln(x) = lim  = lim  =  lim x = 0 0 1/x 1/x^{2}
lim x^{x} = lim e^{x.ln(x)} = e^{lim x.ln(x)} 0 0 and from previous example = e^{0} = 1
lim (cos(x))^{1/x} 0 = lim e^{(1/x).ln(cos(x))} 0 = e^{lim (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} = e^{0} = 1 0
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