Question

Suppose that the differentiable functions $\mathrm{u}, \mathrm{v}, \mathrm{f}\(, \)g: R \rightarrow R\( satisfy \)\lim _{x \rightarrow \infty} u(x)=2, \lim _{x \rightarrow \infty} v(x)=3$ \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty} g(x)=\infty\) and \(\frac{f^{\prime}(x)}{g^{\prime}(x)}+u(x) \frac{f(x)}{g(x)}=v(x)\) then \(\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}\) is equal to (given that it exists) (A) 1 (B) \(1 / 2\) (C) 2 (D) None

Short answer

of the above

Step-by-step solution

Break Down the Given Equation

We have the equation: \(\frac{f^{\prime}(x)}{g^{\prime}(x)}+u(x) \frac{f(x)}{g(x)}=v(x)\). We will first rewrite it to express the term \(\frac{f(x)}{g(x)}\).

Isolate the Term \(\frac{f(x)}{g(x)}\)

Subtract \(\frac{f^{\prime}(x)}{g^{\prime}(x)}\) from both sides to get: \(u(x) \frac{f(x)}{g(x)} = v(x) - \frac{f^{\prime}(x)}{g^{\prime}(x)}\). Now, divide both sides of the equation by u(x) to get: \(\frac{f(x)}{g(x)} = \frac{v(x) - \frac{f^{\prime}(x)}{g^{\prime}(x)}}{u(x)}\).

Find the Limit of LHS

Now, as x approaches infinity, we will compute the limit of the left-hand side (LHS) \(\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}\). To do this, we will compute the limit of the right-hand side (RHS). Since f(x) and g(x) approach infinity and have derivatives given by f'(x) and g'(x), we can apply L'Hôpital's Rule.

Apply L'Hôpital's Rule to RHS

Let's first compute the limit of the RHS: \(\lim _{x \rightarrow \infty} \frac{v(x) - \frac{f^{\prime}(x)}{g^{\prime}(x)}}{u(x)}\). From the given values, \(\lim _{x \rightarrow \infty} v(x) = 3\), and \(\lim _{x \rightarrow \infty} u(x) = 2\). Next, applying L'Hôpital's Rule to the fraction, we get: \(\lim _{x \rightarrow \infty} \frac{\frac{-f^{\prime^2}(x)}{g^{\prime^2}(x)}}{0}\). Let's now assume that this limit exists and is finite. Notice that since \(\lim_{x \rightarrow \infty} f'(x) = \lim_{x \rightarrow \infty} g'(x) = 0\), both numerators approach 0, and it is necessary to apply L'Hôpital's Rule again.

Apply L'Hôpital's Rule Again

After applying L'Hôpital's Rule again, we get: \(\lim _{x \rightarrow \infty} \frac{\frac{f^{\prime^3}(x)}{g^{\prime^3}(x)}}{0}\). Once again, we assume this limit exists and is finite. Finally, since the denominator is approaching 0, the limit of the RHS becomes: \(\lim _{x \rightarrow \infty} \left(\frac{\frac{f^{\prime^3}(x)}{g^{\prime^3}(x)}}{0} \right)= \lim _{x \rightarrow \infty} \left(\frac{f^{\prime^3}(x)}{g^{\prime^3}(x)}\right)\).

Equate LHS and RHS

Now, the limit of LHS is equal to the limit of RHS: \(\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}= \lim _{x \rightarrow \infty} \left(\frac{f^{\prime^3}(x)}{g^{\prime^3}(x)}\right)\).

Evaluate the Answer

After comparing the answer options with the limit of the LHS, we observe that none of the options (1, 1/2, 2) match the answer. Therefore, the correct answer is: (D) None