onegirl
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
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it is a well known limit
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

anonymous
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but the proof is not obvious
look in any intro calculus book, it will be there using a geometric argument and the "squeeze theorem"

onegirl
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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)

anonymous
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1) replace \(x\) by 2

onegirl
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ok

anonymous
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2) factor as
\[\frac{(x3)(x+2)}{x3}=x+2\] then replace \(x\) by 3

onegirl
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okay

anonymous
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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

onegirl
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ok

onegirl
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for number 4) its this one lim t> 2 (1/2 + 1/t over 2 + t)

onegirl
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and 5) lim x>4 + sqrt(16  x^2)

anonymous
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\[\frac{1}{2}+\frac{1}{t}=\frac{t+2}{2t}\] divide by \(t+2\) and get
\[\frac{1}{2t}\] replace \(t\) by \(2\)

onegirl
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ok

anonymous
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\[\lim_{x\to 4^+}\sqrt{16x^2}\] does not exist because if \(x>4\) then \(16x^2<0\) and you cannot take the square root of a negative number

onegirl
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okay thanks alot :)