I am suposed to apply L'Hospitals rule to solve. I can't seem to solve it tho. could someone explain the steps to me? Thanks!
the limit as x approaches infinity of:
xe^(1/x)-x
This is what I got so far:
xe^(1/x)-x
x(e^(1/x)-1)
e^(1/x)-1/(1/x)
Now its in the right form to solve using the rule. so...
e^(1/x)/-1/x^2
But can I simplify it further and how to I now solve as x approaches infinity? I am stuck!
Can you formally prove L'Hôpital's rule, also for this case where you have a zero times infinity case? If not, then I think it is a bad idea to try to do this problem using L'Hôpital's rule.
If you use L'Hôpital's rule, what you are actually doing is saying that near any point x0 a differentiable function behaves like
f(x0+h) = f(x0) + h f'(x0)
And then you take x0 to be the limit point.
Now, in this problem it is easier to start with this more fundamental rule, and say that for small t, we have that:
exp(t) = 1 + t + O(t^2),
where O(t^2) denote a term that is smaller than some constant times t^2.
This means that for large x, we have:
exp(1/x) = 1 + 1/x + O(1/x^2)
And then you obtain:
x [exp(1/x) - 1] = 1 + O(1/x)
The limit for x to infinity is thus 1.
look at the second example of this short video.
It is exactly your question.
(Broken Link Removed)
forgot to mention to click on the 6th video in the list on the left side.
To solve the limit of xe^(1/x) - x as x approaches infinity using L'Hôpital's rule, you have correctly manipulated the expression into the form x(e^(1/x) - 1). Now, you can further simplify it by applying the limit rule to the inner expression (e^(1/x) - 1) and the outer expression (x).
1. Simplifying the inner expression:
e^(1/x) = 1 + 1/x + (1/2!)(1/x)^2 + (1/3!)(1/x)^3 + ... (by the Taylor series expansion)
Therefore,
e^(1/x) - 1
= (1 + 1/x + (1/2!)(1/x)^2 + (1/3!)(1/x)^3 + ...) - 1
= 1/x + (1/2!)(1/x)^2 + (1/3!)(1/x)^3 + ...
2. Now, let's simplify the outer expression:
x
As x approaches infinity, x is growing very large. Therefore, we can ignore the terms with powers of (1/x) because they will approach zero.
Now, the expression simplifies to:
(x(e^(1/x) - 1))/(x)
= e^(1/x) - 1
3. Finally, we can take the limit as x approaches infinity. As x approaches infinity, 1/x approaches 0. So, we have:
lim(x→∞) (e^(1/x) - 1)
= e^0 - 1
= 1 - 1
= 0
Therefore, the limit of xe^(1/x) - x as x approaches infinity is 0.