Question

question 9 (1 point)
let ( f(x) ) be a function such that ( lim _{x
ightarrow 0} f(x)=0 ) and ( lim _{x
ightarrow 0} f^{prime}(x)=-1 ). the limit
lim _{x
ightarrow 0} \frac{f(x) cos (7 x)}{sin (4 x)}
is equal to
( \frac{7}{4} )
0
( \frac{4}{7} )
( -\frac{1}{4} )
( -4 )

Explanation:

Step1: Apply L'Hopital's Rule

Since \(\lim_{x
ightarrow0}f(x) = 0\) and \(\lim_{x
ightarrow0}\sin(4x)=0\), we have \(\frac{0}{0}\) form. By L'Hopital's Rule \(\lim_{x
ightarrow0}\frac{f(x)\cos(7x)}{\sin(4x)}=\lim_{x
ightarrow0}\frac{f^{\prime}(x)\cos(7x)-7f(x)\sin(7x)}{4\cos(4x)}\)

Step2: Evaluate the limit

We know that \(\lim_{x
ightarrow0}f(x) = 0\), \(\lim_{x
ightarrow0}f^{\prime}(x)=- 1\), \(\lim_{x
ightarrow0}\cos(7x)=1\), \(\lim_{x
ightarrow0}\sin(7x)=0\) and \(\lim_{x
ightarrow0}\cos(4x)=1\)
Substitute these values into \(\lim_{x
ightarrow0}\frac{f^{\prime}(x)\cos(7x)-7f(x)\sin(7x)}{4\cos(4x)}\)
We get \(\frac{\lim_{x
ightarrow0}f^{\prime}(x)\cdot\lim_{x
ightarrow0}\cos(7x)-7\lim_{x
ightarrow0}f(x)\cdot\lim_{x
ightarrow0}\sin(7x)}{4\lim_{x
ightarrow0}\cos(4x)}=\frac{-1\times1 - 7\times0\times0}{4\times1}\)

Answer:

\(-\frac{1}{4}\)