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anonymous

  • 5 years ago

limit as x approaches 0 of (sin 5x)/7x?

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  1. JamesJ
    • 5 years ago
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    what's the limit as u --> 0 of (sin u)/u

  2. anonymous
    • 5 years ago
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    1

  3. JamesJ
    • 5 years ago
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    right. So think about trying to rewrite your equation in that form. Hint: the final answer is not 1.

  4. anonymous
    • 5 years ago
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    I am hopeful that I can multiply the numerator by 5x/5x. Then I can cancel out the leftover x in the numerator with the x in the denominator (5x/7x, where the x's cancel ) and have 5/7?

  5. JamesJ
    • 5 years ago
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    yes

  6. anonymous
    • 5 years ago
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    Thanks...this is very humbling...haven't done algebra in years and I'm about a week into a calculus course. You've been very helpful twice now.

  7. JamesJ
    • 5 years ago
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    I don't even remember the first time, but I'm glad I have been. Keep at it. It's fun and intellectually very satisfying.

  8. anonymous
    • 5 years ago
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    I asked about delta-epsilon proofs earlier in the day. I am studying very hard. Thanks again.

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