BRIEF ANSWERS FOR TEST 1

• T or F?
  • F, the limit is zero. (L'Hospital's Rule does not apply.)
  • T. Since f=o(g) we have lim_(x goes to infinity) (f/g) = 0. Therefore f/g is bounded for large x, which implies that f=O(g).
  • F. If x = 0 and C = 1, then this equality asserts that ln(2) = 1. But ln(2) = .697..., and not 1.
  • T, since y = e^x is continuous.
  • F. The limit is +Pi.
• Evaluate ...
  • This limit is in the form 1^(infinity), which is indeterminate. L'Hospital's Rule applies to show that the limit is e^3.
  • If you draw the correct triangle it is easy to see that sec(tan^(-1)(x/2)) = square_root(x^2+4)/2.
• Integrate ...
  • Use the substitution u = tan(x). Then you get an answer of 2^(tan(x))/ln(2) + C.
  • Since 3-2t-t^2 = 2^2 - (t+1)^2, we can use the integration formula involving sin^(-1)(t) that we just learned. We get that (before evaluating at the bounds) the integral is 6(sin^(-1)((t+1)/2)). After evaluating we get just Pi.
• Solve the IVP. Put it into normal form. P(x) = 1/(xln(x)), so nu(x) = ln(x). Thus y(x) = (1/ln(x))[(1/2)ln^2(x) + C]. From y(e)=1 we calculate that C = 1/2. Thus y(x) = (1/2)[ln(x) + 1/ln(x)]. It follows that y(square_root(e)) = 5/4.
• The bounds of integration are -Pi/3 and +Pi/3. The integral of 8cos(x) - sec^2(x) on this interval is 6*square_root(3).
• We need to integrate Pi*sin^2(x) from 0 to Pi. The answer is (Pi^2)/2.
• If you use the arc length formula this reduces to the integral of cosh(x) from 0 to ln(3). The answer is sinh(ln(3)) = 4/3.