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# Calc 1: Practice Exam 2 Solutions Name: (1) Find ... - Westmont

Calc 1: Practice Exam 2 Solutions Name: (1) Find ... - Westmont

## 4(6)

4(6) Find the following limits. Explain all your steps.(a)ln(x)limx→∞ 3√ xAnswer: The given limit has the indeterminate form ∞ ∞so we can use l’Hopital’s rule. We find that the given limitis the same as:limx→∞We can rewrite this as:limx→∞1x13 x−2/33x 1/3Since the denominator goes to +∞ and the numeratorstays at 3 as x goes to ∞ the given limit evaluates to 0.(b)e x − 1 − xlimx→0 x 2Answer: The given limit has the indeterminate form 0 0so we can use l’Hopital’s rule to see that the given limit isequal to:e x − 1limx→0 2xThis limit still has the indeterminate form 0 so we apply0l’Hopital’s rule again:e xlimx→0 2As x → 0 the function e x heads toward 1 and 2 stays at 2,so the overall limit is equal to 1.2(c)limx→∞x 2 − 12x 2 + 1

Answer: We could use l’Hopital’s rule here, but it is simplerto multiply both numerator and denominator of thefraction by 1 converting it into the limit:x 21 − 1 xlim2x→∞ 2 + 1 x 2The function 1 x 2 → 0 as x → ∞ so the overall limit is 1 2 .5(7) Find the point on the line y = −3x + 2 which is closest to theorigin.Answer: Define the function d(x) = √ x 2 + (−3x + 2) 2 whichrepresents the distance of the point (x, y) (where y = −3x + 2)from the origin. We wish to minimize this so we take the derivativeusing the chain rule:d ′ 2x + 2(−3x + 2)(−3)(x) =2 √ x 2 + (−3x + 2) 2Note that the denominator is 2 times d(x) so since the linedoesn’t pass through the origin, the denominator is never zero.Thus the only critical points occur when the numerator is zero.To see when this occurs, solve the following:2x + 2(−3x + 2)(−3) = 0We see that x = 3 5 . So the point nearest the origin is ( 3 5 , 1 5 ).(8) When a person coughs, the velocity of the air stream is relatedto the radius of the trachea by:v(r) = k(r 0 − r)r 2The normal radius of the trachea is r 0 and k is a constant. Theradius r of the trachea is restricted so that 1 2 r 0 ≤ r ≤ r 0 inorder to prevent suffocation.Find the value of r which makes v an absolute maximum.Answer:Rewrite: v(r) = kr 0 r 2 − kr 3 . Then differentiate:v ′ (r) = 2kr 0 r − 3kr 2

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