Question

let ( f ) be the function defined by ( f(x)=2x + 3x^{-10} ), and let ( g ) be a differentiable function with derivative given by ( g(x)=\frac{1}{x}-4cosleft(\frac{1}{x}
ight) ). it is known that ( lim_{x
ightarrowinfty}g(x)=infty ). the value of ( lim_{x
ightarrowinfty}\frac{f(x)}{g(x)} ) is
a 0
b ( \frac{1}{2} )
c 1
d indeterminate

Explanation:

Step1: Apply L - H Rule

Since \(\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}(2x + 3x^{-10})=\infty\) and \(\lim_{x
ightarrow\infty}g(x)=\infty\), by L - H rule \(\lim_{x
ightarrow\infty}\frac{f(x)}{g(x)}=\lim_{x
ightarrow\infty}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Step2: Find \(f^{\prime}(x)\)

Differentiate \(f(x)=2x + 3x^{-10}\) using the power rule \((x^n)^\prime=nx^{n - 1}\). So \(f^{\prime}(x)=2-30x^{-11}\)

Step3: Find \(\lim_{x

ightarrow\infty}f^{\prime}(x)\) and \(\lim_{x
ightarrow\infty}g^{\prime}(x)\)
\(\lim_{x
ightarrow\infty}f^{\prime}(x)=\lim_{x
ightarrow\infty}(2-\frac{30}{x^{11}})=2\)
\(\lim_{x
ightarrow\infty}g^{\prime}(x)=\lim_{x
ightarrow\infty}(\frac{1}{x}-4\cos(\frac{1}{x}))\)
Let \(t = \frac{1}{x}\), as \(x
ightarrow\infty\), \(t
ightarrow0\). Then \(\lim_{t
ightarrow0}(t - 4\cos t)=0 - 4\cos(0)=- 2\)

Step4: Calculate \(\lim_{x

ightarrow\infty}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)
\(\lim_{x
ightarrow\infty}\frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{\lim_{x
ightarrow\infty}f^{\prime}(x)}{\lim_{x
ightarrow\infty}g^{\prime}(x)}=\frac{2}{2}=1\)

Answer:

C. \(1\)