## needhelp07 2 years ago evaluate the limits

1. needhelp07

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

2. shining

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

3. needhelp07

how did u do that??

4. shining

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

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

6. shining

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

7. needhelp07

ahh ok

8. Shanks

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