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4 find the following limits you can use anything we

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4. Find the following limits. (You can use anything we proved in class about limits, but you need to explain your steps!) (a) lim x 0 sin(5 x ) 1 + x - 1 - x = lim x 0 sin(5 x )( 1 + x + 1 - x ) ( 1 + x - 1 - x )( 1 + x + 1 - x ) = lim x 0 sin(5 x )( 1 + x + 1 - x ) 2 x = lim x 0 sin(5 x ) 5 x ( 1 + x + 1 - x ) 2 / 5 The function ( 1+ x + 1 - x ) 2 / 5 is continuous at zero so its limit is ( 1+0+ 1 - 0) 2 / 5 = 5. The limit of sin(5 x ) 5 x is the same as sin( x ) x as x 0 which is 1. Using the fact that the limit of products is the product of limits we get that lim x 0 sin(5 x ) 1 + x - 1 - x = 5 . (b) lim x →∞ 2 3 x + 5 = lim x →∞ 2 /x 3 + 5 /x Since lim x →∞ 1 /x = 0 we have lim x →∞ 2 /x = 0 and lim x →∞ 3 + 5 /x = 3 (using the sum and product rule of limits). Then the limit of the ratio is the ratio of the limits (since the limit of the denominator is not zero), so lim x →∞ 2 3 x +5 = 0. Another way to prove this would be by the application of the squeezing principle. If x > 0 then 0 < 2 3 x + 5 < 2 3 x . Since 1 /x 0 as x → ∞ , this will also be true for 2 3 1 x (by the product rule of limits). Another proof would be by using the ε - δ definition (with a large constant c in place of δ , since we take the limit at ). 2
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4 Find the following limits You can use anything we proved...

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