lim x->0 X*e^(cos 1/x)
The limit is 0.
You can see what a limit will approach by first plugging the number into the function. As x approaches 0, the function will be 0*e^(cos(1/0)). Since there is a 1/0 in the function, we know that the value of the function is undefined at x=0. However, limits don't tell us definite values of a function, but rather how the function behaves over time. To make this easier on us, lets just assume that the first x (before the e) gets smaller and smaller. As the value of x gets smaller from either side, the function's value gets smaller. So long as x does not equal 0, the function will get smaller. You can try graphing this function in your calculator or see a table of values to see the behavior of the function as x approaches 0.
This is also easy to see since cos(1/x) never exceeds 1. So, e^cos(1/x) never exceeds e.
So,
|e^cos(1/x)| <= e
x e^cos(1/x) <= e*x -> 0 as x->0
I agree with SJ but
I would look at the backend of this thing first.
Consider 1/x
As x --->0 , 1/x becomes infinitely large
but we are taking cos(???)
and the result of cos(anything) is a number between -1 and 1
so e^(cos(1/x)) lies between e^-1 and e^1 or between 1/e and e, no matter how big 1/x becomes
Now look at the multiplier of x at the front.
As x gets smaller and smaller, the product will approach 0
Thank you for the help SJ, Steve & Reiny.
To evaluate the limit as x approaches 0 of x*e^(cos(1/x)), we can use the squeeze theorem.
First, let's consider a function g(x) = e^(cos(1/x)). We know that the cosine function oscillates between -1 and 1, so cos(1/x) is bounded between -1 and 1 for all nonzero x. Therefore, we can say 0 <= g(x) <= e for all nonzero x.
Next, let's multiply g(x) by x. We have f(x) = x * g(x). Because x is approaching 0, we have 0 <= x * g(x) <= e * x.
As x approaches 0, we know that x goes to 0, and e * x also goes to 0. Therefore, by the squeeze theorem, the limit of f(x) as x approaches 0 is 0.
In conclusion, the limit as x approaches 0 of x*e^(cos(1/x)) is 0.