Thank you very much for this video. you talk about a sin graph that oscillates back and for the horizontal asymptote but " never quite makes it". what do you mean by that?
You could simply factor out x^2 in the square root in the last example. It becomes x/(sqrt(x^2(9+1/x))+3x) --> x/(x*(sqrt(9+1/x)+3)) --> 1/(sqrt(9+1/x)+3)
![[Calculus] Limits at Infinity Examples](http://i.ytimg.com/vi/9m7PaaO3G-Q/mqdefault.jpg)
![[Calculus] Limits at Infinity Examples](/img/tr.png)
![[Calculus] Limit Practice #1](http://i.ytimg.com/vi/tSA3V_MH4-M/mqdefault.jpg)
Thank you very much for this video.
you talk about a sin graph that oscillates back and for the horizontal asymptote but " never quite makes it". what do you mean by that?
You could simply factor out x^2 in the square root in the last example. It becomes x/(sqrt(x^2(9+1/x))+3x) --> x/(x*(sqrt(9+1/x)+3)) --> 1/(sqrt(9+1/x)+3)
this is another good method to use. just one small mistake; you're missing an x in the second part. it should be x/(x*(sqrt(9+1/x)+3x
Thank you! great explanation!
Could you make a video explaining limits on x*e^(1/x) where x -> 0 from both directions aswell as x -> (+ -)infinity?
Thank you
Thanks
Yeah... I have solved the second part with out seeing the answer