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onegirl
Group Title
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
 one year ago
 one year ago
onegirl Group Title
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
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
it is a well known limit \[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
but the proof is not obvious look in any intro calculus book, it will be there using a geometric argument and the "squeeze theorem"
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
1) 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)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
1) replace \(x\) by 2
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
2) factor as \[\frac{(x3)(x+2)}{x3}=x+2\] then replace \(x\) by 3
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
3) 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
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
for number 4) its this one lim t> 2 (1/2 + 1/t over 2 + t)
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
and 5) lim x>4 + sqrt(16  x^2)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{2}+\frac{1}{t}=\frac{t+2}{2t}\] divide by \(t+2\) and get \[\frac{1}{2t}\] replace \(t\) by \(2\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\lim_{x\to 4^+}\sqrt{16x^2}\] does not exist because if \(x>4\) then \(16x^2<0\) and you cannot take the square root of a negative number
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
okay thanks alot :)
 one year ago
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