b approaches 0 from below, so 1/b tends to minus infinity. There is a small mistake: the 1 in the (1 . 0 - 0) term. Need to use the fact that x e-x tends to 0 as x tends to +infinity.Phys12 said:
In this solution, in the last 3rd line, I get the first part (-e^-1 - e^-1), however, after the '-' symbol, the person writes (1/b * e^1/b - e^1/b) and takes the limit as b->0. However, shouldn't this give him (inf. * e^inf - e^inf)?
Thanks
(His answer is correct, by the way)
Ohk! I didn't notice the fact that we were approaching 0 from the left hand side. Thanks! :)haruspex said:
Phys12 said:
The limit of e^(1/x) as x approaches 0 is equal to infinity. This can be seen by taking the limit of the function as x approaches 0 from both the positive and negative sides, which will result in a very large positive value.
We can prove this by using the definition of a limit, which states that for a function f(x), the limit as x approaches a is equal to L if for any positive number ε, there exists a positive number δ such that if 0 < |x-a| < δ, then |f(x) - L| < ε. In the case of e^(1/x) as x approaches 0, we can choose a value for δ that will result in a very large value for f(x), thus showing that the limit is infinity.
No, the limit of e^(1/x) as x approaches 0 can only be positive infinity. This is because as x approaches 0 from the positive side, e^(1/x) will approach infinity, and as x approaches 0 from the negative side, e^(1/x) will approach 0. Since infinity is not a finite number, it cannot have a negative counterpart.
Yes, it is possible for the limit of e^(1/x) as x approaches 0 to not exist. This can happen if the function oscillates or if the limit approaches different values from the positive and negative sides.
The graph of e^(1/x) as x approaches 0 is a vertical asymptote at x=0. As x gets closer and closer to 0, the graph will get steeper and steeper, approaching infinity on the positive side and 0 on the negative side.