Question

evaluate the limits

Answer 1

\[\lim_{x \rightarrow \infty}(\frac{ x ^{3}-1 }{ x ^{4}+1 })\]

Answer 2

easily consider the highest rate of varaiabe seperately in nominator and denominator
lim x^3/x^4 = 0

Answer 3

how did u do that??

Answer 4

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

Answer 5

what happens to -1 and +1 ?? is it cancelled out?? sorry if i have many questions

Answer 6

you shoul eliminate them because in the limits of fraction like this type constants have no touchable influence

Answer 7

ahh ok

Answer 8

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