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anonymous
 3 years ago
Evaluate the indicated limit, if it exists. If it does not exist, explain why it doesn’t. Assume that lim x> 0 sin x/x = 1
anonymous
 3 years ago
Evaluate the indicated limit, if it exists. If it does not exist, explain why it doesn’t. Assume that lim x> 0 sin x/x = 1

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is a well known limit \[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the proof is not obvious look in any intro calculus book, it will be there using a geometric argument and the "squeeze theorem"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01) lim x> 2 (x5/x^2 + 4) 2) lim x> 3 (x^2  x  6/x  3) 3) lim h> 0 ((2 + h)^2  4/h)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01) replace \(x\) by 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02) factor as \[\frac{(x3)(x+2)}{x3}=x+2\] then replace \(x\) by 3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.03) expand get \[\frac{(2+h)^24}{h}=\frac{4+4h+h^24}{h}=\frac{4h+h^2}{h}=4+h\] then replace \(h\) by 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for number 4) its this one lim t> 2 (1/2 + 1/t over 2 + t)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and 5) lim x>4 + sqrt(16  x^2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}+\frac{1}{t}=\frac{t+2}{2t}\] divide by \(t+2\) and get \[\frac{1}{2t}\] replace \(t\) by \(2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to 4^+}\sqrt{16x^2}\] does not exist because if \(x>4\) then \(16x^2<0\) and you cannot take the square root of a negative number
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