What is the limit of cosx/log(x-phi/2+1) if lim x is to phi/2
we have 0/0, so try
-sinx / (1/(x-pi/2+1))
= -1 / (1/1)
= -1
To find the limit of cos(x) / log(x - φ/2 + 1) as x approaches φ/2, we can use L'Hôpital's rule, which states that the limit of the ratio of two functions that both approach zero or infinity can be found by taking the derivative of the numerator and denominator successively, until a determinable limit is reached.
First, let's differentiate the numerator and the denominator separately.
The derivative of cos(x) is -sin(x), and the derivative of log(x - φ/2 + 1) is 1 / (x - φ/2 + 1).
Now, we can calculate the limit of the ratio of these derivatives as x approaches φ/2:
lim x -> φ/2 (-sin(x) / (1 / (x - φ/2 + 1)))
Simplifying this expression, we get:
lim x -> φ/2 -sin(x) * (x - φ/2 + 1)
Next, we can directly substitute φ/2 into this expression:
-sin(φ/2) * (φ/2 - φ/2 + 1) = -sin(φ/2)
Therefore, the limit of cos(x) / log(x - φ/2 + 1) as x approaches φ/2 is -sin(φ/2).