Calculus Help (squeeze Theorem)?

I think I know the first question and im not really sure where to start with the second. Was hoping someone could walk me through the steps. I couldn't find any examples online that were setup like the second. I appreciate any help.

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  • Anonymous
    3 years ago

    If you find these hard, it means you haven’t understood what the squeeze theorem mean. It’s actually very simple. Try one of the many YouTube videos.

    Let f(x) = 1 -x³/3 and let g(x) = 1 – x

    Lim[x→3]f(x) = 1 – 3³/3 = 1 – 3 = -2

    Lim[x→3]g(x) = 1 – 3 = -2

    f(x) ≤ u(x) ≤ g(x) tells you that u(x) is always bigger or equal to f(x), and u(x) is always smaller or equal to g(x). u(x) is ‘sandwiched’ between f(x) and g(x).

    That means Lim[x→3]u(x) = -2

    (Any other value for u(x) would mean u(x) is not ‘sandwiched’ between f(x) and g(x).)

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    You could take the same approach for the 2nd question.

    Let f(x) = 3 and g(x) = x+4

    Lim [x→-1] of f(x) = 3 (because f(x) is always 3!)

    Lim [x→-1] of g(x) = -1+4 = 3

    Since f(x) ≤ j(x) ≤ g(x), that means Lim [x→-1]j(x) = 3. j(x) is sandwiched between 3 and 3.

    Lim[x→-1](1+j(x))/x = (1 + 3)/-1 = -4

  • 3 years ago

    1 - ((x^2)/3) <= u(x) <= 1 - x

    If x = 3, we have:

    1 - ((3^2)/3) <= u(x) <= 1 - 3

    1 - (9/3) <= u(x) <= -2

    1 - 3 <= u(x) <= -2

    -2 <= u(x) <= -2

    u(x) = -2

    3 <= j(x) <= x + 4

    If x = -1, we have:

    3 <= j(x) <= -1 + 4

    3 <= j(x) <= 3

    j(x) = 3

    (1 + j(x)) / (-1) = (1 + 3) / (-1) = 4/(-1) = -4

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