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Unformatted text preview: Math 408C Practice Final Problems Key 1 Limits Evaluate the following limits. 1. lim x → π 2 x 2 π 4 √ x π Factoring the numerator, we observe that lim x → π 2 x 2 π 4 √ x π = lim x → π 2 ( x + π 2 )( √ x + π )( √ x π ) √ x π = lim x → π 2 ( x + π 2 )( √ x + π ) = ( π 2 + π 2 )( π + π ) = 4 π 3 . 2. lim x → π  x π  x π For all x < π we note that  x π  = π x , so lim x → π  x π  x π = lim x → π π x x π = lim x → π 1 = 1 . 3. lim x → π +  x π  x π For all x > π we note that  x π  = x π , so lim x → π +  x π  x π = lim x → π + x π x π = lim x → π + 1 = 1 . 1 4. lim x → π x 3 ( x π ) 5 We observe that lim x → π x 3 ( x π ) 5 =∞ and lim x → π + x 3 ( x π ) 5 = + ∞ . Therefore the limit does not exist. 5. lim x → sin( πx ) x lim x → sin( πx ) x = lim x → π sin( πx ) πx = π lim x → sin( πx ) πx = π. 6. lim x →∞ sin( πx ) x Since 1 /x ≤ sin( πx ) /x ≤ 1 /x for x > 0, the Squeeze Theorem gives that lim x →∞ 1 x ≤ lim x →∞ sin( πx ) x ≤ lim x →∞ 1 x . Since lim x →∞ 1 x = lim x →∞ 1 x = 0 , it follows that lim x →∞ sin( πx ) x = 0 . 7. lim x →∞ x π 3 Because π 3 > 0, lim x →∞ x π 3 = ∞ . 2 8. lim x →∞ x π 4 Because π 4 < 0, lim x →∞ x π 4 = 0 . 9. lim x →∞ πx 3 x 3 + x 2 + x + 1 Dividing numerator and denominator by x 3 , we find lim x →∞ πx 3 x 3 + x 2 + x + 1 = lim x →∞ π 1 + 1 /x + 1 /x 2 + 1 /x 3 = lim x →∞ π lim x →∞ 1 + lim x →∞ (1 /x ) + lim x →∞ (1 /x 2 ) + lim x →∞ (1 /x 3 ) = π 1 + 0 + 0 + 0 = π. 10. lim x →∞ cos( π x ) As x → ∞ , π x → 0. Since cos is a continuous function, it follows that lim x →∞ cos( π x ) = lim a → cos( a ) = cos(0) = 1 . 3 2 Differentiation 11. If d dx f ( x ) = 0, what can be said about f ( x )? If d dx f ( x ) = 0, this means the slope of f ( x ) is zero everywhere. Since this means f never increases nor decreases, f must be constant. Compute dy/dx for the following function using the limit definition of the derivative. 12. y = x 3 + x . d dx [ x 3 + x ] = lim h → [( x + h ) 3 + ( x + h )] [ x 3 + x ] h = lim h → x 3 + 3 x 2 h + 3 xh 2 + h 3 + x + h x 3 x h = lim h → 3 x 2 h + 3 xh 2 + h 3 + h h = lim h → (3 x 2 + 3 xh + h 2 + 1) = 3 x 2 + 1 . Compute dy/dx for the following relations. 13. y = sec x ln x By the quotient rule, dy dx = (ln x )(tan x sec x ) (sec x )(1 /x ) (ln x ) 2 4 14. y = π x x π By the product rule, dy dx = ( π x ln π )( x π ) + ( π x )( πx π 1 ) 15. y = p 1 /x dy dx = d dx x 1 / 2 = 1 2 x 3 / 2 . 16. y = (tan x sec x )(tan x + sec x ) Solution 1: We could do this with the product rule: dy dx = (tan x sec x )(sec 2 x + tan x sec x ) + (sec 2 x tan x sec x )(tan x + sec x ) = (tan x sec x )(sec x + tan x ) sec x + sec x (sec x tan x )(tan x + sec x ) = 0 ....
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This note was uploaded on 09/18/2011 for the course M 408 D taught by Professor Textbookanswers during the Fall '07 term at University of Texas.
 Fall '07
 TextbookAnswers
 Factoring, Limits

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