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SOLUTIONS TO HOMEWORK ASSIGNMENT #1 1. Compute the following limits: (3x2 + 2x + 1)10 (a) lim (x3 − 3x2 + 5) (b) lim (c) lim (−3t3 − 4t + 5)1/3 x→−2 x→−1 (x3 + 5)5 t→2 √ t2 − 9 t − 9t 3 3− z (d) lim (e) lim (f) lim t→3 t − 3 t→−3 t + 3 z→9 9 − z SOLUTIONS: (a) lim (x3 − 3x2 + 5) = (−2)3 − 3(−2)2 + 5 = −15 by substituting x = −2 and using x→−2 continuity of the polynomial x3 − 3x2 + 5 at x = −2. (3x2 + 2x + 1)10 (3(−1)2 + 2(−1) + 1)10 210 (b) lim = = 5 = 1 by substituting x = −1 and x→−1 (x3 + 5)5 ((−1)3 + 5)5 4 2 10 (3x + 2x + 1) using continuity of the function at x = −1. (x3 + 5)5 (c) lim (−3t3 − 4t + 5)1/3 = (−3(2)3 − 4(2) + 5)1/3 = (−27)1/3 = −3 by substituting t = 2 t→2 and using continuity of the function (−3t3 − 4t + 5)1/3 at t = 2. t2 − 9 (t − 3)(t + 3) (d) lim = lim = lim(t + 3) = 6. Notice that we can not put t = 3 right t→3 t − 3 t→3 t−3 t→3 0 away as it leads to the nonsense . This is why we must ﬁrst do some algebra, namely 0 cancel t − 3 from both denominator and numerator. Once this is done we can evaluate the limit by putting t = 3 and using continuity of the linear function t + 3. t3 − 9t t(t − 3)(t + 3) (e) lim = lim = lim t(t − 3) = 18. t→−3 t + 3 t→−3 t+3 t→−3 √ √ √ 3− z 3− z 3+ z 9−z 1 1 (f) lim = lim × √ = lim √ = lim √ = . No- z→9 9 − z z→9 9−z 3+ z z→9 (9 − z)(3 + z) z→9 3 + z 6 0 tice that we could not directly set z = 9 as this leads to the nonsense , but after some 0 1 algebra it became posible. Setting z = 9 in the last step is valid since the function √ 3+ z is continuous at z = 9. 2. Find an equation of the tangent line of y = f (x) at x = a for the following: x+1 √ (a) y = x2 + x, a = 2 (b) y = , a = 3 (c) y = x + 1, a = 3 x−1 SOLUTIONS: The equation of the tangent line to the graph of y = f (x) at x = a is always given by y = f (a) + f (a)(x − a). We use the laws of diﬀerentiation to compute the derivatives. 1 (a) If f (x) = x2 + x then f (x) = 2x + 1, and so the equation of the tangent line is y = 6 + 5(x − 2). x+1 −2 (b) If f (x) = then f (x) = , and so the equation of the tangent line is x−1 (x − 1)2 1 y = 2 − (x − 3). 2 √ 1 (c) If f (x) = x + 1 then f (x) = √ , and so the equation of the tangent line is 2 x+1 1 y = 2 + (x − 3). 4 3. Evaluate the following limits: √ 1 1 x+4−2 (a) lim √ −1 (b) lim h→0 h 1+h x→0 x SOLUTIONS: (a) √ √ √ 1 1 1 1− 1+h 1 1− 1+h 1+ 1+h lim √ −1 = lim √ = lim × √ × √ h→0 h 1+h h→0 h 1+h h→0 h 1+h 1+ 1+h 1 1 − (1 + h) −1 = lim ×√ √ = lim √ √ h→0 h 1 + h(1 + 1 + h) h→0 1 + h(1 + 1 + h) 1 = − 2 (b) √ √ √ x+4−2 x+4−2 x+4+2 (x + 4) − 4 lim = lim ×√ = lim √ x→0 x x→0 x x+4+2 x→0 x( x + 4 + 2) x 1 1 = lim √ = lim √ = x→0 x( x + 4 + 2) x→0 x+4+2 4 4. A certain function y = f (x) satisﬁes f (1) = −1, f (1) = 2. (a) Determine an equation for the tangent line at x = 1. (b) Find the x and y intercepts of the tangent line. (c) Graph the tangent line. SOLUTIONS: (a) The equation is y = f (1) + f (1)(x − 1), that is y = −1 + 2(x − 1). (b) The x and y intercepts are x = 3/2 and y = −3 respectively. 2 (c) 1 x –1 –0.5 0.5 1 1.5 2 0 –1 –2 –3 –4 –5 Figure 1: The tangent line of y = f (x) at x = 1 3

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posted: | 6/22/2010 |

language: | English |

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