MATH114 Lecture Notes - Lecture 10: Squeeze Theorem, Algebraic Solution, Asymptote

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ultramarineeel54

27 Sep 2018

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University of Alberta

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Mathematics

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MATH114

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Powell,Elizabeth

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MATH114 Lecture 9: MATH114-LECTURE#9-LIMITS OF A FUNCTION

MATH114 Lecture Notes - Lecture 10: Squeeze Theorem, Algebraic Solution, Asymptote

MATH114 Lecture Notes - Lecture 11: Indeterminate Form, Intermediate Value Theorem

Document Summary

Math114-lecture#10-limits of a function continued and squeeze theorem. If f(x) grows without bound as x goes to a, we write: lim (cid:1858)(cid:4666)(cid:1876)(cid:4667)= and say that f approaches infinity as x approaches a. For example, find lim (cid:2868)(cid:2869)2 looking at the value it approaches, Algebraically, (cid:2869)(cid:2868) dne, (cid:271)ut sin(cid:272)e we"re f(x) approaches as x approaches 0 zero from the left, f(x) approaches , right, f(x) approaches , too. Looking the graph: as x approaches and as x approaches zero from the. If f(x) decreases without bound as x goes to a, we write: lim (cid:1858)(cid:4666)(cid:1876)(cid:4667)= and say that f approaches negative infinity as x goes to a. Using the same concept as the one used in the example above, the value f(x) approaches as x approaches zero. However, in this function it is (cid:2869)(cid:2868) thus f(x) approaches graph approaches from the left lim (cid:2868) (cid:2869) (cid:4666) (cid:2870)(cid:4667)2= and the right side and since the left.

Related Questions

Show transcribed image text This Question: 1 pt 12 of 44 (0 complete) This Quiz: 44 pts possible Verify Property 2 of the definition of a probability density function over the given interval. ffx) 22x What is Property 2 of the definition of a probability density function? O A. The area under the graph of f over the interval [a,bl is b O B. The area under the graph of f over the interval [a.b] is 1 O C. The area under the graph of f over the interval [a,b is a. Since the given interval. [0.oo), has a right-endpoint of infinity, the area under the graph of fwill be calculated using an improper integral. Choose the correct formula for an improper integral over the interval [a,ool below t(x) dx= lim | f(x) dx= lim [F(x)]a= lim (F(b)-F(a)) b-o0 f(x) dx = lim | f(x) dx = lim [F(x)]-lim (F(a)-F(b)) b- 00 b--0o f(x) dk- lim x) dx- lim [F(x)m (F(b)-F(a)) b-+00
O D. 00 f(x) dx= lim f(x) dx= lim [F(æ£´= lim (F(a)-F(b)) b--oã Substitute a and b into the left side of the formula from the previous step and express the integral as a limit. area 2edx lim2edx Next, determine F(x). First, find the antiderivative of f. 2(Use C as the artbitrary constant.)
Next, determine F(x). First, find the antiderivative of f. as the anbliay cosat) Let C 0 in the expression obtained above and let the resulting expression be F(x). Evaluate the result over [0,oo) using the far right side of the formula for the area. area = As b approaches oo, e approaches infinity where c is a constant. Use this fact to calculate the limits from the previous step area+1 Is Property 2 of the definition of a probability density function over the given interval now verified? Choose the correct answer below A Property 2 o the definition o a probability density unction over the given interval has been ver ed since the expression in the previous step simplifies to o B. Property 2 of the definition of a probability density function over the given interval has not been verified because the expression in the previous step does not simplify to the expected area value. Â° C. â D Property 2 ofthe definition of a probability density function over the given interval has been verified since the expression in the previous step simplifies to 0 Property 2 o the definition o a probability density unction over the given interval has been ve ned since the expressio n e evious ste si plifies to 1.

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3. (21 pts) Evaluate the limit or say that the limit does not exist. If a limit does not exist, explain why. Show your work. You may NOT use a table of values, a graph, or L'Hospitals's Rule to justify your answer. (Note: Possible answers include + or -..) (a) (7 pts) x2 - 2x - 3 - lim x+3 Vx+1 - 2 x2 - 2x - 3 Vx+ 1 + 2 lim x+3 Vx+1 -2 Vx+1 +2 (x - 3) (x + 1) (Vx+ = lim x+3 x + 1 - 4 = lim x+3 (x - 3) (x + 1) (Vx+1 +2). -- x-3 = lin (x + 1) (Vx+1 +2) = (3 + 2) (V3+1 +2) = = 4 (V4 +2) = 16 (b) (7 pts) lim xsin (In x) X >0+ NOTE: We cannot apply the Product Law, since lim sin (1n x) DOES NOT EXIST. On the other hand, -1s sin (in x) s 1 and , therefore, -xs x sin (in x) sx, (since x > 0). Since lim X = lim (-X) = 0, x0+ X0 the inequality above implies that lim xsin (in x) = 0, by the Squeeze Theorem . x0+ EXPLANATION : Since lim ln x = -0, x →0+ sin (lnx) oscillates between 1 and -1 as x + 0*, and, therefore lim sin (In x) DOES NOT EXIST. x >0+ (c) (7 pts) in x+2- (lnx) +In 20 -= -00 (x - 2) 70-

MATH114 Lecture Notes - Lecture 10: Squeeze Theorem, Algebraic Solution, Asymptote

MATH114 Lecture 9: MATH114-LECTURE#9-LIMITS OF A FUNCTION

MATH114 Lecture Notes - Lecture 10: Squeeze Theorem, Algebraic Solution, Asymptote

MATH114 Lecture Notes - Lecture 11: Indeterminate Form, Intermediate Value Theorem

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