Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 63

Given the function lim_{x->0^+}(4x+1)^cot(x)
We have to find the limit.
The given function is an intermediate form of type 1^infty which can be written as:
i.e lim_{x->0^+}f(x)^g(x)=e^(lim_{x->0^+}g(x)[f(x)-1])
e^(lim_{x->0^+}g(x)[f(x)-1])=e^(lim_{x->0^+}cot(x)[4x+1-1])=e^(lim_{x->0^+}cot(x)4x)
=e^(lim_{x->0^+}(4cos(x).x)/sin(x))
=e^(lim_{x->0^+}(4cos(x))/(sin(x)/x))
We know that lim_{x->0}sin(x)/x=1
Therefore we get,
e^(lim_{x->0^+}4cos(x))=4
Therefore the limit is e^4.