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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 xØ¶ x 2 b. Lim ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ xØ¶ x+10 x3 -2 x c. Lim ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ ex xØ¶ x d.Lim ÅÅÅÅÅÅÅÅ ÅÅÅ xØ ¶ lnHxL 4 x2 +8 x-100 e. Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ xØ¶ x2 -2 x+3 -3 x+1 -3 f. Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ 2 xØ¶ 2 x+3 1 ÅÅÅÅÅÅ g. Lim ÅÅÅÅ ÅxÅÅÅÅÅ xØ¶ 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 xØ¶ 1 LnI1+ ÅÅÅÅÅ M x = Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 ÅÅÅÅÅÅ xØ¶ 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 xØ¶ LnI1+ ÅÅ1ÅÅÅ M Form ÅÅÅÅ00 x = Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 ÅÅÅÅÅÅ ÅÅÅÅÅ x xØ¶ L' Hopital = 1 -1 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ * ÅÅÅÅÅÅÅÅÅÅ 1 2 1+ ÅÅÅÅÅ x x ÅÅÅÅÅÅÅÅÅ Lim ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ -1 xØ¶ ÅÅÅÅÅ2ÅÅÅÅÅ x 1 = Lim ÅÅÅÅÅÅÅÅ 1ÅÅÅÅÅ xØ¶ 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 xØ¶ 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