## anonymous 3 years ago evaluate the limits

1. anonymous

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

2. anonymous

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

3. anonymous

how did u do that??

4. anonymous

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. anonymous

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

6. anonymous

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

7. anonymous

ahh ok

8. anonymous

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