Question about taking limits in calculus
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Question about taking limits in calculus

[From: ] [author: ] [Date: 11-04-29] [Hit: ]
now whats infinity minus infinity? 0hope i helped!!!......

2) im gonna address the bottom function here first because again, it just cancels to 1. so now the top function. the -1/4 goes away because against its deriv is 0. now 1/x is equal to x^-1 correct? so its derivative would be -x^-2 or -1/x^2. plugging in 4 here we get -1/4^2 which is -1/16

3) this one is definitely the trickiest. ok, so you're gonna expand your equation to be one giant squareroot. its gonna be the sqrt of your top things squared over 4x^2+5. that ends up being 4x^2-12x+9 over 4x^2+5. the limit as x--> thing can now actually go inside the radical and you can start evaluating from there. use L'Hopitals rule to take the deriv of the top, 8x-12, over the deriv of the bottom 8x. that just simplified to 1-(3/2x). now the limit as x--> thingy is gonna jump again the inside of the parenthesis around 3/2x. the limits goes the bottom of the fraction and you evaluate 3 divided by the limits of 2x as x goes to infinity. this ends up being 0. now its just the sqrt of 1-0 which is 1. phew

4) this one is actually incredibly simple to evaluate just by looking at it. what is the limit as x approaches infinity? its infinity. what the limit as sqrt(x^2+4) approaches infinity? well its also infinity seeing as infinity-squared + 4 is still gonna be infinity because its this inconceivably unreachable number. now whats infinity minus infinity? 0

hope i helped!!!
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