wesdg1978
  • wesdg1978
lim x->0 (x^2-2x+sinx/x) How do I find the limit using: lim x=0 sinx/x=1
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hartnn
  • hartnn
just separate out the limits, lim x->0 (x^2-2x+sinx/x) = lim x->0 (x^2) - lim x->0 (2x) +lim x->0 (sinx/x) the last limit is =1 for others, just plug in x=0
wesdg1978
  • wesdg1978
So the answer is simply 1?
hartnn
  • hartnn
or was it (x^2-2x+sin x) / x ? if its (x^2-2x+sinx/x) then yes, limit =1

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wesdg1978
  • wesdg1978
|dw:1371418962260:dw|
hartnn
  • hartnn
|dw:1371419073685:dw|
hartnn
  • hartnn
now its not just 1
wesdg1978
  • wesdg1978
That sux, 'cause now I'm just as lost as to begin with
hartnn
  • hartnn
why u lost, you still separate, you still plug in x=0 lim x->0 (x^2-2x+sinx/x) = lim x->0 (x^2/x) - lim x->0 (2x/x) +lim x->0 (sinx/x)
wesdg1978
  • wesdg1978
This is what I see: 0-0+1=1 Is that wrong?
hartnn
  • hartnn
yeah 2x/x = 2 0-2+1 got this ?
wesdg1978
  • wesdg1978
Cancel the x's before plugging in 0 for x, I just overlooked that part. Trying to over complicate it I guess. Thanks for your help!
hartnn
  • hartnn
welcome ^_^

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