Topics Midterm 1 Review Worksheet Sections 1 5 1 6 1 7 1 8 2 1 2 2 2 3 2 4 2 5 Any errors you can nd in the solutions can be reported here and are greatly appreciated https forms gle rGXwBeet5a3c3kF6A Many solutions here will be quite brief since the methods are the same as in previous worksheets On exams however the more work you can show neatly the better 1 Evaluate the limit if it exists If the limit does not exist state whether it is or neither You may NOT use L H opitals rule Show your work If you use a theorem clearly state which theorem you are using sin 4 sin 5 a lim 0 2 Partial solution Rewrite the function as Now we can see the limit is 20 cid 18 4u cid 19 6u4 b lim u 0 u6 sin Partial solution Squeeze theorem Since limu 0 u 6 0 then limu 0 u6 sin cid 18 1 x4 c lim x 0 1 1 x4 x4 cid 19 sin 4 sin 5 20 4 5 cid 18 4u 6u4 cid 19 cid 12 cid 12 cid 12 cid 12 u 6 cid 12 cid 12 cid 12 cid 12 u6 sin cid 19 cid 18 4u 0 0 6u4 cid 19 cid 18 1 x4 x4 lim x 0 lim x 0 1 1 x4 Solution Find a common denominator then multiply by conjugate 1 x4 1 x4 1 x4 1 x4 1 1 x4 x4 x4 x4 1 x4 1 1 x4 x4 1 x4 1 lim x 0 lim x 0 lim x 0 1 x4 1 1 x4 1 1 x4 1 1 x4 1 1 x4 1 x4 1 lim x 0 1 2 3x 15 x 5 4 if x cid 54 5 if x 5 d lim x 5 g x where g x Partial solution Factor 3x 15 The answer is 3 t2 4t 7 x 5 3 x 5 x 5 3 when x cid 54 5 The function value g 5 4 is a distraction e t 6 lim t 6 Solution The numerator approaches 62 4 6 7 53 and the denominator approaches 0 over a range where t 6 is negative The limit does not exist it diverges to 1 f lim t 3 4t 12 t 3 Partial solution Write t 3 separately We nd cid 40 t 3 t 3 t 3 t 3 4t 12 t 3 lim t 3 lim t 3 t 3 4 Therefore limt 3 4t 12 t 3 4t 12 lim t 3 4 t 3 t 3 lim t 3 4 4 4t 12 t 3 does not exist and similarly limt 3 x2 4cos 1 x3 3 x 2 g lim x 0 Now evaluate the left hand and right hand limits Partial solution Split up the limit as a sum Use the squeeze theorem to show the rst limit is equal to 0 via the inequalities 1 cos x 3 1 4 1 4cos x 3 41 x24 1 x24cos x 3 x241 Since the red and blue terms approach 0 as x 0 the Squeeze Theorem says limx 0 x24cos x 3 0 tan 4 h lim 0 8 limit Partial solution Write the tangent function in terms of sine and cosine and use similar techniques to the Sept 30 worksheet 2 Is there a number a such that limx 5 exists If so nd the value of a and the value of the x2 x ax 2a 1 x2 3x 10 Solution The denominator factors as x2 3x 10 x 5 x 2 Therefore the limit exists when x 5 is a factor of the numerator which happens exactly when x 5 is a root of the numerator which means The limit exists when a 3 Plug in a 3 yourself and compute the limit 52 5 5a 2a 1 0 a 3 2 3 Consider the function f x x2 2x 3 x2 1 a Identify all values of x for which the function is discontinuous Solution We can see this just by factoring the denominator x2 1 x 1 x 1 The function is continuous where ever it is de ned and it is de ned everywhere except x 1 and x 1 b Does the function have a removable discontinuity If so for which value s of x c Does the function have a jump discontinuity If so for which value s of x d For any removable discontinuities write down the corresponding continuous extension Solution Now let s factor more thoroughly f x x 3 x 1 x 1 x 1 There is a hole removable discontinuity at x 1 and an asymptote at x 1 There are no jump discontinuities Rational functions only have vertical asymptotes and removable discontinuities We can compute We should de ne f 1 2 to remove the removable discontinuity lim x 1 f x lim x 1 x 3 x 1 1 3 1 2 2 3 4 Does the function f x sin x 2x 4 have a root on the interval 0 2 Explain why If you use a theorem clearly state which theorem you are using Solution Yes f 0 4 and f 2 4 4 0 Since 0 lies between 4 and 4 4 therefore f x 0 for some intermediate 0 x 2 by the intermediate value theorem 5 Find a number such that if 0 x 5 then 4x 20 cid 15 where cid 15 0 1 Solution Therefore 1 40 is suitable 6 Let f x 2x 11 We know that lim x 1 0 x 1 Solution f x 9 Given cid 15 0 3 nd 0 such that f x 9 cid 15 when Therefore 3 20 is a suitable value 7 Find the values of a and b that make f continuous everywhere Partial solution The only dangers of non continuity are at x 1 and x 2 We need to ensure and The rst equations reduce to The second equations reduce to Solving this system of equations yields a b 2 f x is continuous 3 1 3 Our work shows that when a b 2 3 1 3 we ensure 4x 20 0 1 4 x 5 0 1 x 5 1 40 f x 9 cid 15 2x 11 9 0 3 3 10 3 20 2x 2 x 1 f x ax2 bx 2x a bx x2 4 x 2 if x 1 if 1 x 2 if x 2 lim x 1 f x lim x 1 f x f 1 x 2 f x lim lim x 2 f x f 2 a b 2 a b a b 1 4 a 2b 4 a 2b 0 4 8 Suppose f x is continuous on 0 x 7 and the only solutions of the equation f x 3 are x 1 and x 6 If f 4 2 then which of the following options is correct a f 2 3 b f 2 3 c it is not possible to …
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