question 9 (1 point)\nlet ( f(x) ) be a function such that ( lim _{x \rightarrow 0} f(x)=0 ) and ( lim _{x…

question 9 (1 point)\nlet ( f(x) ) be a function such that ( lim _{x \rightarrow 0} f(x)=0 ) and ( lim _{x \rightarrow 0} f^{prime}(x)=-1 ). the limit\nlim _{x \rightarrow 0} \frac{f(x) cos (7 x)}{sin (4 x)}\nis equal to\n( \frac{7}{4} )\n0\n( \frac{4}{7} )\n( -\frac{1}{4} )\n( -4 )

question 9 (1 point)\nlet ( f(x) ) be a function such that ( lim _{x \rightarrow 0} f(x)=0 ) and ( lim _{x \rightarrow 0} f^{prime}(x)=-1 ). the limit\nlim _{x \rightarrow 0} \frac{f(x) cos (7 x)}{sin (4 x)}\nis equal to\n( \frac{7}{4} )\n0\n( \frac{4}{7} )\n( -\frac{1}{4} )\n( -4 )

Answer

Explanation:

Step1: Apply L'Hopital's Rule

Since (\lim_{x\rightarrow0}f(x) = 0) and (\lim_{x\rightarrow0}\sin(4x)=0), we have (\frac{0}{0}) form. By L'Hopital's Rule (\lim_{x\rightarrow0}\frac{f(x)\cos(7x)}{\sin(4x)}=\lim_{x\rightarrow0}\frac{f^{\prime}(x)\cos(7x)-7f(x)\sin(7x)}{4\cos(4x)})

Step2: Evaluate the limit

We know that (\lim_{x\rightarrow0}f(x) = 0), (\lim_{x\rightarrow0}f^{\prime}(x)=- 1), (\lim_{x\rightarrow0}\cos(7x)=1), (\lim_{x\rightarrow0}\sin(7x)=0) and (\lim_{x\rightarrow0}\cos(4x)=1) Substitute these values into (\lim_{x\rightarrow0}\frac{f^{\prime}(x)\cos(7x)-7f(x)\sin(7x)}{4\cos(4x)}) We get (\frac{\lim_{x\rightarrow0}f^{\prime}(x)\cdot\lim_{x\rightarrow0}\cos(7x)-7\lim_{x\rightarrow0}f(x)\cdot\lim_{x\rightarrow0}\sin(7x)}{4\lim_{x\rightarrow0}\cos(4x)}=\frac{-1\times1 - 7\times0\times0}{4\times1})

Answer:

(-\frac{1}{4})