a problem of limits: lim, x - >infty, sin((x^2 + 1)/(x + 1))
- Thread starter blackhat
- Start date
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
Re: a problem of limits
-1<sin(x^2+1)<1, so as x+1 goes to infinity, the numerator oscilates between -1 and 1, hence the limit has to be zero.
Note: This is not divergence by oscillation. the limit of sin(x) as x approaches infinity is undefined as sin(x) oscillates between -1 and 1,no limit.
Note: lim of 1/x, -1/x as x approaches infinity, f(x) = ±1/x approaches zero.
-1<sin(x^2+1)<1, so as x+1 goes to infinity, the numerator oscilates between -1 and 1, hence the limit has to be zero.
Note: This is not divergence by oscillation. the limit of sin(x) as x approaches infinity is undefined as sin(x) oscillates between -1 and 1,no limit.
Note: lim of 1/x, -1/x as x approaches infinity, f(x) = ±1/x approaches zero.
- Joined
- Feb 4, 2004
- Messages
- 16,582
You have the sine of a rational expression. From what you learned, back in algebra, about asymptotes, you know that the argument of the sine has a slant asymptote of y = x - 1. Obviously, this "goes to infinity" as x "goes to infinity".blackhat said:
Since the argument value does not have a finite limit, then the sine wave will never "settle down"; instead, as mentioned in one of the previous replies, it will continue to oscillate back and forth between -1 and +1.
As such, there can be no defined limit value. :wink: