University:
Boston University
Course:
CGS MA 121 | Calculus
Academic year:

2024
Views:

168

Pages:

3
Author:

fiziokrat16l

Question 1 Find the following limits or state that they do not exist. lim h→0 3 / √(16 + 3h) + 4 Answer Evaluate: lim h→0 3 / √(16 + 3h) + 4 Consider the given expression: lim h→0 3 / √(16 + 3h) + 4 = 3 / √(16 + 3 * 0) + 4 = 3 / √16 + 4 = 3 / (4 + 4) = 3/8 Hence, lim h→0 3 / √(16 + 3h) + 4 = 3/8 Question 2 Use L'Hospital Rule to evaluate the following limits. lim x→0 (tan hx)^2 Answer Consider the given expression, lim x→0 (tan hx)^2 Since the function does not have any denominator hence, the L'Hospital rule cannot be applied. Thus, lim x→0 (tan hx)² = lim x→0 (e^(hx) - e^(-hx) / (e^(hx) + e^(-hx)))² = lim x→0 (e^(0) - e^(0) / (e^(0) + e^(0)))² = (1 - 1 / (1 + 1))² = (0/2)² = 0 Which is the required value. Question 3 For what values of p is each series convergent? Σ n=1→∞ ((-1)^(n-1)) / n^p Answer Σ n=1→∞ ((-1)^(n-1)) / n^p This is an alternating series, but for p < 0 we can try the Test for Divergence first to see if we can rule out convergence quickly before trying the Alternating Series Test. If p<0 then we get lim n→∞ ((-1)^(n-1)) * n^|p| = DNE It is oscillating and doesn't converge Question 4 Evaluating trig limit lim x→0 (√(1 - cos x²)) / (1 - cos x) I have tried to simplify the expression using the identity 1 - cos x = 2sin²(x/2), but I have still failed to remove the indeterminate form. Answer 1 Hint: 1 - cos α = 2sin²(α/2) Apply this with both numerator and denominator. Then use lim x→0 sin x / x = 1 lim x→0 x / sin x = 1 Answer 2 A little more details: lim x→0 √(1 - cos(x²)) / (1 - cos(x)) = lim x→0 √(2sin²(x²/2)) / (2sin²(x/2)) = lim x→0 (√2 |sin(x²/2)|) / (2sin²(x/2)) = √2 lim x→0 |sin(x²/2)| / (2sin²(x/2)) = √2 lim x→0 |sin(x²/2)| / 2(sin(x/2))² = √2 lim x→0 |sin(x²/2)|/ (x²/2) * (x²/2) / 2(sin(x/2))² = √2 lim x→0 |sin(x²/2)| / (x²/2) * (x/2)² / 2(sin(x/2))² * x⁴ / x² = √2 lim x→0 (sin(x²/2)) / (x²/2) * x⁴ / x² * (x²/4) / 2(sin²(x/2)) = √2 * 1 * lim x→0 (x⁴ / 4x²) = √2 lim x→0 (x²/4) = √2 * 0 = 0 Question 5 The limit as x approaches zero of (sin x - xcos x) / x³ Answer Let f(x) = g(x) / h(x) , where g(x) = sin x - x cos x , and h(x) = x³. The limit is x → 0. Since g(0) and h(0) both = 0, then use l'Hôpital's rule, lim x→0 f(x) = lim x→0 g'(x) / h'(x) g'(x) = cos x - (cos x - x sin x) = x sin x h'(x) = 3x² Since g'(0) and h'(0) both = 0, then again use l'Hôpital's rule, i.e. lim x→0 f(x) = lim x→0 g''(x) / h''(x) g''(x) = sin x + x cos x h''(x) = 6x Since g''(0) and h''(0) both = 0, then again use l'Hopital's rule, i.e. lim x→0 f(x) = lim x→0 g'''(x) / h'''(x) g'''(x) = cos x + cos x - x sin x = 2 cos x - x sin x h'''(x) = 6 Now, g'''(0) = 2, and h'''(0) = 6. So, lim x→0 f(x) = lim x→0 g'''(x) / h'''(x) = 2 / 6 = 1/3