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onegirl

  • 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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  1. anonymous
    • 3 years ago
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    it is a well known limit \[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

  2. anonymous
    • 3 years ago
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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"

  3. onegirl
    • 3 years ago
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    1) lim x-> 2 (x-5/x^2 + 4) 2) lim x-> 3 (x^2 - x - 6/x - 3) 3) lim h-> 0 ((2 + h)^2 - 4/h)

  4. anonymous
    • 3 years ago
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    1) replace \(x\) by 2

  5. onegirl
    • 3 years ago
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    ok

  6. anonymous
    • 3 years ago
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    2) factor as \[\frac{(x-3)(x+2)}{x-3}=x+2\] then replace \(x\) by 3

  7. onegirl
    • 3 years ago
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    okay

  8. anonymous
    • 3 years ago
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    3) expand get \[\frac{(2+h)^2-4}{h}=\frac{4+4h+h^2-4}{h}=\frac{4h+h^2}{h}=4+h\] then replace \(h\) by 0

  9. onegirl
    • 3 years ago
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    ok

  10. onegirl
    • 3 years ago
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    for number 4) its this one lim t-> -2 (1/2 + 1/t over 2 + t)

  11. onegirl
    • 3 years ago
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    and 5) lim x->4 + sqrt(16 - x^2)

  12. anonymous
    • 3 years ago
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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 \(-2\)

  13. onegirl
    • 3 years ago
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    ok

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

  15. onegirl
    • 3 years ago
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    okay thanks alot :)

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