We are tasked with evaluating the limit:
L=x→0limcscx(2cos2x+3cosx−cos2x+sinx+4) Substituting
x=0 yields an indeterminate form of
∞⋅0. To resolve this, we rewrite the expression as a fraction:
L=x→0limsinx2cos2x+3cosx−cos2x+sinx+4 To eliminate the radicals, we multiply the numerator and the denominator by the conjugate expression:
C(x)=2cos2x+3cosx+cos2x+sinx+4 Applying the identity
(a−b)(a+b)=a2−b2, the numerator becomes:
(2cos2x+3cosx)−(cos2x+sinx+4)
Distributing the negative sign and simplifying, we obtain:
cos2x+3cosx−sinx−4
We recognize the quadratic component
cos2x+3cosx−4. By substituting
u=cosx, we factor the expression as
(u+4)(u−1), which gives:
(cosx+4)(cosx−1)−sinx
The limit expression now takes the form:
L=x→0lim[sinx⋅C(x)(cosx+4)(cosx−1)−sinx⋅C(x)sinx]
For the first term, we utilize the identity
sinxcosx−1=−tan(2x). As
x→0, this term approaches
0:
x→0limC(x)(cosx+4)(−tan(x/2))=255⋅0=0 Substituting
x=0 into the denominator, we get
5+5=25. Thus, the final result is: