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needhelp07

  • 3 years ago

evaluate the limits

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  1. needhelp07
    • 3 years ago
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    \[\lim_{x \rightarrow \infty}(\frac{ x ^{3}-1 }{ x ^{4}+1 })\]

  2. shining
    • 3 years ago
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    easily consider the highest rate of varaiabe seperately in nominator and denominator lim x^3/x^4 = 0

  3. needhelp07
    • 3 years ago
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    how did u do that??

  4. shining
    • 3 years ago
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    i said you should just put the highest power of varaiable instead of up and down (because the limit approaches infinity) so\[\lim \frac{ x^{3} }{ x^{4}? }\]=0 because x^4 > x^3

  5. needhelp07
    • 3 years ago
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    what happens to -1 and +1 ?? is it cancelled out?? sorry if i have many questions

  6. shining
    • 3 years ago
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    you shoul eliminate them because in the limits of fraction like this type constants have no touchable influence

  7. needhelp07
    • 3 years ago
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    ahh ok

  8. Shanks
    • 3 years ago
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    you can understand this by taking x^2 and x^4 common as follows: => \[\lim_{x \rightarrow \infty} \left( \frac{ x^{2} ( 1- \frac{ 1 }{x^{2}}) }{ x^{4} ( 1+ \frac{ 1 }{x^{4}}) } \right)\] Further strike out and reduce it to \[\lim_{x \rightarrow \infty} \left( \frac{( 1- \frac{ 1 }{x^{2}}) }{ x^{2} ( 1+ \frac{ 1 }{x^{4}}) } \right)\] Obviously 1/x^2 and 1/x^4 when x-> infinity will be = 0 So inside bracket fraction becomes 1/1. and 1/x^2 = 0 when x-> infinity Thereby, given lim = 0

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