Evaluate the limit if it exists, otherwise, write ∞,-∞ or DNE as appropriate. (you may not use the L'Hospital Rule)
lim e^cosx \squareroot(tan((3)/(4)x)+10)
x-> π
Please show steps I am very confused!
So am I. If you mean
e^cosx * √(tan(3/4 x)+10)
then I see no problem. tan(3/4 π) = -1
cos π = -1
so the expression at x = π is just
e^-1 √(-1+10) = 3/e
Not sure why a limit is even involved, as there are no 0/0 or ∞ involved anywhere
If I read it wrong, maybe you could format things as I did, and clarify
To evaluate the limit of the given function, we will substitute π for x in the function and simplify the expression as much as possible. Here are the steps:
1. Substitute π for x in the function:
lim (x->π) e^cosx / √(tan((3x)/(4))+10)
2. Evaluate the functions at x = π:
e^cos(π) / √(tan((3π)/(4))+10)
3. Simplify the expression:
e^(-1) / √(tan((3π)/(4))+10)
Now, let's simplify the expression further by evaluating the remaining trigonometric function and calculate the limit.
The expression tan((3π)/(4)) can be simplified as follows:
tan((3π)/(4)) = tan(π - π/4) = -tan(π/4)
The value of tan(π/4) is 1, so now we can substitute this value into the expression:
e^(-1) / √(-tan(π/4) + 10)
Now, we have:
e^(-1) / √(-1 + 10) = e^(-1) / √(9) = e^(-1) / 3
Therefore, the limit of the function as x approaches π is e^(-1) / 3.