Download LINEAR APPROXIMATION, LIMITS, AND L`HOPITAL`S RULE v.03

29LHopital.nb
1
LINEAR APPROXIMATION, LIMITS, AND
L'HOPITAL'S RULE v.03
ü Who was L'Hopital. Taken from Catholic Enciclopedia at
http://www.newadvent.org/cathen/07469a.htm
Guillaume-François-Antoine de L'Hôpital
Marquis de Sainte-Mesme and Comte d'Entremont, French mathematician; b. at Paris, 1661; d. at Paris, 2 February, 1704.
Being the son of the lieutenant-general of the king's armies he was intended for a military career and served for some time as
captain in a cavalry regiment. He had no talent for Latin but early displayed extraordinary ability for mathematics. At the age
of fifteen he had solved a number of problems proposed by Pascal, and while an army officer, he studied mathematics in his
tent. Owing to extreme near sightedness he was forced to resign and then devoted himself entirely to his favourite studies. In
1692 he became acquainted with Jean Bernoulli one of the three or four men of the day who understood the new methods of
differential calculus. During four months he studied with Bernoulli whom he had invited to his estate of Oucques near
Vendôme and learned from him this branch of the science of numbers. In 1693 he was elected honorary member of the
Academy of Sciences of Paris and soon rivalled Newton Huyghens Leibniz and the Bernoullis in the propounding and
solving of problems involving the calculus. He is remembered because he made it possible for others to learn this new
system. His work on the analysis of the infinitesimal for the study of curves was published in 1696 and was received with
great satisfaction by many who were trying to solve the mystery surrounding these advanced problems for the book contained
a clear and careful exposition of the methods employed. The rule for the evaluation of a fraction whose numerator and
denominator both have a limit value of zero is named after L'Hôpital. His wife is said to have been associated with him in his
work. His published works are : "Analyse des infiniment petits pour l'intélligence des lignes courbes" (Paris, 1696; last ed. by
Lefèvre, Paris, 1781); "Traité anlytique des sections coniques" (Paris, 1707; 2nd ed., 1720)several memoirs and notes
inserted in the "Recueil de l'Académie des sciences" (Paris, 1699-1701), and in "Acta Eruditorum" (Leipzig, 1693-1699).
INTRODUCTION
¶
Some times we are dealing with limits of functions of the form ÅÅÅÅ0 or ÅÅÅÅ
Å . We can calculate some of these limits using the
0
¶
sequence approach but not the general properties we discussed to find the limit of a function. However, we will be able to
calculate these limits using L'Hopital's rule. This rule is based on the linear approximation of a function at a point. As an
application we will study more carefully the dominance of functions when x goes to plus or minus infinity, and in general the
end behavior of functions.
Linear Approximation
Nearby a point at which a function is differentiable, the function and its tangent line are approximately the same. The tangent
line at that point on the curve is called the linearization of the curve on the neighborhood of the point.
EXAMPLE 1
Consider the function f HxL = x2 nearby x = 1. The equation of the tangent line at this point is y - f H1L = f ' H1L Hx - 1Lor
y = 2 Hx - 1L + 1. The graph below shows the graph of the function and its derivative in a window centered at x = 1.If you
change the size of the window to shrink it around x = 1, you will see that the tangent line approximates the function better.
29LHopital.nb
2
EXAMPLE 1
Consider the function f HxL = x2 nearby x = 1. The equation of the tangent line at this point is y - f H1L = f ' H1L Hx - 1Lor
y = 2 Hx - 1L + 1. The graph below shows the graph of the function and its derivative in a window centered at x = 1.If you
change the size of the window to shrink it around x = 1, you will see that the tangent line approximates the function better.
0
Zoom of function at x = 1
1
2
1
1
0
1
2
EXERCISE 1
a. What is the concavity of the function at x = 1? Justify your answer using the second derivative.
b. When the linear approximation at x = 1 is used to estimate the function for values in a neighborhood of 1, are those
values undersestimates or overestimates? Why?
c. Use the linear approximation to estimate the value of y = x2 at x = 1.01 and x = 0.98
d. How good was the estimate? In other words, what is the error in the estimate? Look at the graph and then verify it
algebraically.
e. You are asked to estimate the value of the function at x = 2 using hte linearization at x=1. Is it going to be a good
estimate? Justify your answer.
EXERCISE 2
a. Find the linearization of y = 2-x at x = -2.
b. Are the values obtained with this linearization and overestimate/underestimate? Justify your answer graphically and using
derivatives.
c. What is the maximum error estimating the value of function one can made using its linearization at x = -2on the interval
@-2.2, -1.8D? Justify your answer. The graphic may help to determine it.
4.6
4.4
4.2
4
3.8
3.6
-2.2
-2.1
-2
-1.9
d. On which neighborhood of -2 the error using the linearization will be less than 0.001?
-1.8
EXERCISE 2
a. Find the linearization of y = 2-x at x = -2.
b. Are the values obtained with this linearization and overestimate/underestimate? Justify your answer graphically and using 3
29LHopital.nb
derivatives.
c. What is the maximum error estimating the value of function one can made using its linearization at x = -2on the interval
@-2.2, -1.8D? Justify your answer. The graphic may help to determine it.
4.6
4.4
4.2
4
3.8
3.6
-2.2
-2.1
-2
-1.9
-1.8
d. On which neighborhood of -2 the error using the linearization will be less than 0.001?
A function f HxL that is differentiable at x = a can be approximated locally by a linear function. This function corresponds to
the tangent line to the function at x = a, and has equation
f HxL @ f £ HaL Hx - aL + f HaL
When we say that the function can be approximated locally it means that on a neighborhood of a we can estimate the value
of the function using its linearization. There in a error involved on in the case that the function is not linear.
EXERCISE 3
è!!!!!!!!!!
Find the linear approximation of y = x - 1 at x = 3. Use this to estimate the function at x = 3.2. Is this an overestimate?
underestimate? Why?
29LHopital.nb
4
The error made with the linearization
è!!!!!!!!!!
Let's consider the linear approximation to y = x - 1 at x = 5to understand what we mean by the error. The linear approximation is y = ÅÅÅÅ1 x + ÅÅÅÅ3 . We want to know when we can use the linear approximation and being accurate in our calculation within
4
4
0.05. It is, the difference between the actual value and the value of the linear approximation is less than 0.05, or the distance
between the function and the linear approximation is less than 0.05. To find those values we solve
è!!!!!!!!!!
dI ÅÅÅÅ14 x + ÅÅÅÅ34 , x - 1 M < 0.05
è!!!!!!!!!!
… ÅÅÅÅ14 x + ÅÅÅÅ34 - x - 1 … < 0.05
è!!!!!!!!!!
-0.05 < ÅÅÅÅ14 x + ÅÅÅÅ34 - x - 1 < 0.05
è!!!!!!!!!!
è!!!!!!!!!!
x - 1 - 0.05 < ÅÅÅÅ14 x + ÅÅÅÅ34 < x - 1 + 0.05
Graphically, these are the points for which the line ÅÅÅÅ14 x + ÅÅÅÅ34 is between the graph of
è!!!!!!!!!!
è!!!!!!!!!!
y = x - 1 - 0.05 and y = x - 1 + 0.05. In the same graph will see the the graphs of
è!!!!!!!!!!
è!!!!!!!!!!
è!!!!!!!!!!
1
3
y = ÅÅÅÅ4 x + ÅÅÅÅ4 , y = x - 1 , y = x - 1 - 0.05 and y = x - 1 + 0.05.
2.4
2.2
2
1.8
1.6
1.4
3
4
5
6
7
è!!!!!!!!!!
From the graph we can see that since the function is concave down, the tangent line will intersect y = x - 1 + 0.05. So the
è!!!!!!!!!!
y = x - 1 + 0.05
solution is given by the solution to 9
y = ÅÅÅÅ1 x + ÅÅÅÅ3
4
4
We can use the Solve command to obtain the interval [3.41115, 6.98885]. This means that if we take any value on this
interval and evaluate the original function or its linearization at that point the maximum error would be 0.05
è!!!!!!!!!!
1
3
SolveA x - 1 + 0.05 == ÅÅÅ
Å x + ÅÅÅ
Å , xE
88x Ø 3.41115<, 8x Ø 6.98885<<
4
4
EXERCISE 4
a. Find the linearization of y = ‰x+1 at x = -2.
b. Determine the interval for which the error made using the linearization is less than 0.001
29LHopital.nb
5
THE LIMIT OF A QUOTIENT OF FUNCTIONS
QUOTIENT RULE FOR LIMITS: We learned that the limit of the quotient of two functions can be calculated as
Lim f HxL
f HxL
xØa
Lim ÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅ as long as both limits exist and the limit of the denominator is not zero.
gHxL
Lim gHxL
xØa
xØa
EXERCISE 5
Calculate the limits below if you can use the quotient rule. If you can not use the quotient rule indicate so.
2
x -3
a. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x+4
xØ2
sinH x L
b. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ
x+1
xØ0
2
x -4
c. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x-2
xØ2
1
d. Lim ÅÅÅÅ
ÅÅÅÅ
»x»
xØ0
29LHopital.nb
6
EXPRESSIONS FOR WHICH WE CAN NOT USE THE QUOTIENT RULE FOR
LIMITS
The following limits can not be calculated using the quotient rule mentioned above. Try them and you will see it.
x-1
2
-1
a. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
ÅÅ
x2 - 1
xØ1
sinHxL
b. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x
xØ0
sinHx-1L
c. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅ
x-1
xØ1
LnHxL
d. Lim ÅÅÅÅÅÅÅÅ
1 ÅÅÅÅÅ
xØ0+
ÅÅÅÅÅÅ
x
¶
The first three functions are of the form ÅÅÅÅ0 , while the last one is of the form ÅÅÅÅ
Å.
0
¶
0
We calculate the limits of the form ÅÅÅÅ0 through their linear approximation at the point in consideration as long as the functions
are differentiable at that point.
f HxL
2
-1
Let's look at the the first limit. This can be written as ÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
ÅÅ . Nearby x = 1 we use their linear approximations to
gHxL
x2 - 1
x-1
obtain:
f HxL
Lim ÅÅÅÅÅÅÅÅ
Å º
gHxL
xØ1
f H1L Hx-1L+ f H1L
Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
g' H1L Hx-1L+gH1L
'
xØ1
Using the linear representation nearby x = 1
An important fact is that f H1L = gH1L = 0
f H1L Hx-1L+0
≈ Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅ
g' H1L Hx-1L+0
'
f H1L Hx-1L
≈ Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
g' H1L Hx-1L
xØ1
'
f H1L
f H1L
Ln2
≈ Lim ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅÅÅÅÅ
Å
g' H1L
g' H1L
2
xØ1
'
'
xØ1
Important to observe three things:
1. f H1L = gH1L = 0, it is the function intercepts the x - axisat x = 1. This fact allowed us to write the tangent lines as
f £ H1L Hx - 1L and g £ H1L Hx - 1L.
2. We can cancell Hx - 1L from numerator and denominator since x Ø 1but it never is 1, reducing the expression inside the
f £ H1L
limit to ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
g£ H1L
3. g'(1)≠0 which allows us to calculate the limit since we don't have any undetermination in the denominator.
f £ H1L
These facts are saying that nearby x = 1, the quotient ÅÅÅÅf HxL
ÅÅÅÅÅÅ º ÅÅÅÅÅÅÅÅ
£ ÅÅÅÅÅÅ , which is the main idea behind L'Hopital's rule.
gHxL
g H1L
The graph below shows the functions together with their linearizations.
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
2. We can cancell Hx - 1L from numerator and denominator since x Ø 1but it never is 1, reducing the expression inside the
f £ H1L
limit to ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
g£ H1L
3. g'(1)≠0 which allows us to calculate the limit since we don't have any undetermination in the denominator.
29LHopital.nb
f HxL
f £ H1L
These facts are saying that nearby x = 1, the quotient ÅÅÅÅ
ÅÅÅÅÅÅ º ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ , which is the main idea behind L'Hopital's rule.
gHxL
g £ H1L
7
The graph below shows the functions together with their linearizations.
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
0.7
0.8
0.9
1
1.1
1.2
L'Hopital's rule
VERSION I: Quotients of the form ÅÅÅÅ00 .
If f HxL and gHxL are differentiable at x = a satisfying:
1. f HaL = gHaL = 0,
2. g£ HaL ≠ 0, then
f HxL
f HaL
Lim ÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
gHxL
g £ HaL
£
xØa
In a simpler language it says that to calculate the limit of the quotient of two functions at a point a, where the limit is of
the form ÅÅÅÅ00 , it is the same as the limit of the quotient of their derivatives at that point.
EXERCISE 6
Consider the functions
sinHxL
a. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x
logHx-1L
b. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅ
x-1
xØ0
xØ1
‰ Verify that the conditions to apply L'Hopital's rule are met.
‰ Evaluate each of the limits by hand using L'Hopital's rule.
¶
VERSION II: Quotients of the form ÅÅÅÅ
Å.
¶
If f HxL and gHxL are differentiable at x = a satisfying:
1. Lim f HaL = Lim gHaL = ± ¶,
2. g£ HaL ≠ 0, then
xØa
xØa
f HxL
f HaL
Lim ÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
gHxL
g £ HaL
£
xØa
f HxL
gHxL
1
This is a consequence of Version I, since if ÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ Ø 0
1 ÅÅÅÅÅ and when f HxL Ø ¶, ÅÅÅÅ
gHxL
f HxL
1
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
f HxL
1.3
If f HxL and gHxL are differentiable at x = a satisfying:
1. Lim f HaL = Lim gHaL = ± ¶,
xØa
xØa
29LHopital.nb
2. g£ HaL ≠ 0, then
8
f HaL
Lim ÅÅÅÅf HxL
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
£ ÅÅÅÅÅ
£
xØa
g HaL
gHxL
gHxL
1
This is a consequence of Version I, since if ÅÅÅÅf HxL
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
1 ÅÅÅÅÅ and when f HxL Ø ¶, ÅÅÅÅÅÅÅÅÅÅ Ø 0
1ÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
gHxL
f HxL
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
f HxL
Note: L'Hopital's rule also holds if we replace the limiting point a for ¶, -¶, a+ , a- .
Some times we need to use L'Hopital's rule more than once. In that case we will use a generalized form of L'Hopital's
rule
f HxL
f £ HxL
Lim ÅÅÅÅ
ÅÅÅÅÅ = Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
gHxL
g £ HxL
xØa
xØa
EXERCISE 7
Calculate the following limits.
x+1
a. Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ
x2 -2 x+3
xض
x
2
b. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
xض x+10
x3 -2 x
c. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ
ex
xض
x
d.Lim ÅÅÅÅÅÅÅÅ
ÅÅÅ
xØ ¶ lnHxL
4 x2 +8 x-100
e. Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
xض x2 -2 x+3
-3 x+1
-3
f. Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ
ÅÅÅÅ
2
xض 2 x+3
1
ÅÅÅÅÅÅ
g. Lim ÅÅÅÅ
ÅxÅÅÅÅÅ
xض 2 -x
Other Limit Forms when L'Hopital's rule can be applied
¶
There are some limits in which we can use L'Hopital's rule after proper changes are made to obtain the form either ÅÅÅÅ0 or ÅÅÅÅ
ÅÅ .
0
¶
These are some cases:
1. Form 0 * •.
LnHxL
¶
x
For example Lim xLnHxL. This limit can be written as Lim ÅÅÅÅÅÅÅÅ
1 ÅÅÅÅÅ , which is of the form ÅÅÅÅÅ , or Lim ÅÅÅÅÅÅÅÅ
1ÅÅÅÅÅÅÅÅ , which is
xØ0+
xØ0 +
form ÅÅÅÅ0 . After that use L'Hopital's rule.
of the
ÅÅÅÅÅÅ
x
¶
ÅÅÅÅÅÅÅÅÅÅÅ
xØ0+ ÅÅÅÅÅÅÅÅ
LnHxL
0
2. Form 00 .
For example Lim xx . First of all bring the exponent to multiply using logarithms.
Lim LnHxx L = Lim x * LnHxL
xØ0+
xØ0+
Form 0 * ¶
xØ0 +
¶
Form ÅÅÅÅ
Å
LnHxL
= Lim ÅÅÅÅÅÅÅÅ
1 ÅÅÅÅÅ
xØ0 +
¶
ÅÅÅÅÅ
x
1
ÅÅÅÅÅ
x ÅÅÅÅÅ
= Lim ÅÅÅÅÅÅÅÅ
1
xØ0 + - ÅÅÅÅÅ2ÅÅÅÅÅ
x
= Lim H-xL = 0
xØ0 +
Since we applied Ln to the original expression, we must apply ‰ to the answer to recover the original one. Hence,
Lim x x
Lim xx = ‰xØ0+
xØ0+
= ‰0 = 1
3. Form 1• .
x
Consider LimI1 + ÅÅÅÅ1 M
x
xØ ¶
Since the variable appear as exponent we should apply Ln to bring it down, to obtain
Lim x LnI1 + ÅÅÅÅ1 M
Form ¶ * 0
x
xض
1
LnI1+ ÅÅÅÅÅ M
x
= Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
1 ÅÅÅÅÅÅ
xض
L' Hopital
= Lim
ÅÅÅÅÅ
x
-1
ÅÅÅÅÅÅÅÅ1ÅÅÅÅÅÅÅÅÅÅ * ÅÅÅÅ
ÅÅÅÅÅÅ
1
2
1+ ÅÅÅÅÅ x
x ÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
-1
Form ÅÅÅÅ0
0
Lim x x
Lim xx = ‰xØ0+
xØ0+
= ‰0 = 1
29LHopital.nb
3. Form 1• .
x
Consider LimI1 + ÅÅÅÅ1x M
9
xØ ¶
Since the variable appear as exponent we should apply Ln to bring it down, to obtain
Lim x LnI1 + ÅÅÅÅ1x M
Form ¶ * 0
xض
LnI1+ ÅÅ1ÅÅÅ M
Form ÅÅÅÅ00
x
= Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
1 ÅÅÅÅÅÅ
ÅÅÅÅÅ
x
xض
L' Hopital
=
1
-1
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ * ÅÅÅÅÅÅÅÅÅÅ
1
2
1+ ÅÅÅÅÅ x
x ÅÅÅÅÅÅÅÅÅ
Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
-1
xض
ÅÅÅÅÅ2ÅÅÅÅÅ
x
1
= Lim ÅÅÅÅÅÅÅÅ
1ÅÅÅÅÅ
xض 1+ ÅÅÅÅÅÅ
x
=1
To recover the original expression we have:
1 x
Lim LnAI1+ ÅÅÅÅÅ M E
x
LimI1 + ÅÅÅÅ1x M = ‰ xØ ¶
x
xØ ¶
= ‰1 = ‰
4. Form •0 .
x
Find Lim I ÅÅÅÅ1x M . This limit is of the form ¶0
xØ0+
After you apply Ln to both sides you obtain a limit of the form 0*¶
x
LnA Lim I ÅÅÅÅ1x M E = Lim x * LnI ÅÅÅÅ1x M
of the form 0 * ¶
xØ0 +
xØ 0+
1Å M
LnI ÅÅÅÅ
= Lim ÅÅÅÅÅÅÅÅ1ÅÅÅÅx ÅÅÅÅ
xØ 0+
ÅÅÅÅÅ
x
x*I ÅÅ1ÅÅÅ M
I ÅÅÅÅÅÅ M
x
£
= Lim ÅÅÅÅÅÅÅÅ1ÅxÅÅÅ£ÅÅÅÅÅ
xØ 0+
¶
of the form ÅÅÅÅ
Å
¶
applying L ' Hopital ' s
=0
To recover the original limit we proceed as before :
x
LnA Lim I ÅÅ1ÅÅÅ M E
x
xØ0 +
Lim I ÅÅÅÅ1x M = ‰
x
xØ0+
= ‰0 = 1
EXERCISE 8
¶
Identify the form of the following limits I ÅÅÅÅ0 , ÅÅÅÅ
ÅÅ , 0 * ¶, 00 M and then proceed to evaluate each of them.
0
¶
x
1. Lim x
xØ0+
1
2. Lim x ÅÅxÅÅÅÅ
xض
3. Lim x2 ‰x
4. Lim H-LnHxLLx
xØ-¶
xØ0+
29LHopital.nb
10
APPLICATION
2
x -2 x
Consider the graph of the function y = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ . Determine:
‰2 x
1. Domain and intercepts with the coordinate axis.
2. Critical points
3. Local maxima and local minima.
4. Concavities of the function.
5. End behavior.
6. Produce a window where we can see a "complete" graph of the function.
EXERCISE 9
Find the limits below. Use L'Hopital's rule where appropiate. If there is a simpler method to solve the problem, use it. If
there are cases when you can not use L'Hopital's indicate why not.
x+1
a. Lim ÅÅÅÅ
ÅÅÅÅÅ
x-1
xØ -1
3
x -5
b. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x4 -1
xØ 1
sinHxL
c. Lim ÅÅÅÅÅÅÅÅ
3ÅÅÅÅÅ
xØ 0
x
LnHLnHxLL
d. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x
xØ ¶
e
e. Lim ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
xØ 2 x-2
è!!!
f. Lim x LnHxL
x+3
g. LimI ÅÅÅÅ14ÅÅÅ - ÅÅÅÅ12ÅÅÅ M
xØ 0 +
1
1
h. LimI ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ - ÅÅÅÅ
ÅÅÅÅÅ M
LnHxL
x-1
xØ 0
x
x
xØ 1
i. Lim xsinHxL
px
j. Lim Hx - 1L TanI ÅÅÅÅ
ÅÅÅÅ M
2
xØ 0
k. LimI ÅÅÅÅxÅÅÅÅÅ M
xØ 1 +
xØ ¶ x+1
x