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anonymous

  • 4 years ago

estimate the limit by substituting smaller and smaller values of h, use radians. h---0 cos(h)-1/h I know you are suppos to use mumbers like .01 .001 .1 etc to then find a common limity, jsut confused on the rad part. thank you

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  1. TuringTest
    • 4 years ago
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    Your calculator probably has at least two ways to write angles: degrees and radians\[360^{\circ}=2\pi \text{radians}\]You really should know about radians in general if you have taken trigonometry, but as far as the question is concerned just make sure your calculator is in radian mode and not degrees.

  2. TuringTest
    • 4 years ago
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    ...radians are another unit for angles besides degrees basically (a much more mathematically logical unit to use as well!)

  3. anonymous
    • 4 years ago
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    yeah i got ya on that, i was just confused if I should use like a radian measure that I know like cos(1pi/6 ) equaling sqrt3/2 or what not to get it to be .001

  4. anonymous
    • 4 years ago
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    i jsut dont know what rad to get to equal .001

  5. TuringTest
    • 4 years ago
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    the question is probably not asking you to find when the answer is 0.001, that doesn't help you much with the limit you are probably just plugging in h=0.001

  6. anonymous
    • 4 years ago
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    ok i got ya ill try that

  7. TuringTest
    • 4 years ago
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    ...but if you wanted to know for what value of h we had the function equal to 0.001 you would solve\[\cos h-\frac1 h=0.001\]which I can't really think how to do offhand

  8. TuringTest
    • 4 years ago
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    ...there isn't much sense to that though

  9. campbell_st
    • 4 years ago
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    the limit will tend to be undefined.... because as you substitute smaller values of h 1/h becomes extemely large.... look at the problem in 2 parts.... cos(h) will approach 1....

  10. anonymous
    • 4 years ago
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    ^^ correct plugged it into the graph calculator. It is 0 or undefined. Thank you for the help guys!

  11. TuringTest
    • 4 years ago
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    the limit should never be zero, it should be \[-\infty\]or undefined as campbell said. welcome!

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