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Find a value for a constant to make a given limit finite and non-zero

by RoRi

Define two functions as follows:

    \[ f(x) = \int_0^x e^{2t} (3t^2 + 1)^{\frac{1}{2}} \, dt, \qquad g(x) = x^c e^{2x}. \]

Find a value for the constant c so that the limit

    \[ \lim_{x \to +\infty} \frac{f'(x)}{g'(x)} \]

is finite and non-zero. Compute the value of the limit.

First, we find the derivatives f'(x) and g'(x),

    \begin{align*} && f(x) &= \int_0^x e^{2t} (3t^2 + 1)^{\frac{1}{2}} \, dt \\[9pt] \implies && f'(x) &= e^{2x} (3x^2+1)^{\frac{1}{2}} \\[9pt] && g(x) &= x^c e^{2x} \\[9pt] \implies && g'(x) &= 2x^c e^{2x} + cx^{c-1} e^{2x} \\[9pt] &&&= e^{2x} (2x^c + cx^{c-1} ). \end{align*}

Then, we evaluate the requested limit,

    \begin{align*} \lim_{x \to +\infty} \frac{f'(x)}{g'(x)} &= \lim_{x \to +\infty} \frac{ e^{2x} (3x^2+1)^{\frac{1}{2}}}{e^{2x} (2x^c + cx^{c-1})} \\[9pt] &= \lim_{x \to +\infty} \frac{(3x^2+1)^{\frac{1}{2}}}{2x^c + cx^{c-1}}. \end{align*}

Now, we would like this limit to be finite and non-zero. To find a value of c that makes this happen, let’s try to apply L’Hopital’s,

    \begin{align*} \lim_{x \to +\infty} \frac{(3x^2+1)^{\frac{1}{2}} }{2x^c + cx^{c-1}} &= \lim_{x \to +\infty} \frac{ \frac{3x}{(3x^2+1)^{\frac{1}{2}}}}{2cx^{c-1} + c(c-1)x^{c-2}} \\[9pt] &= \lim_{x \to +\infty} \frac{ \frac{3}{ \left( 3 + \frac{1}{x^2} \right)^{\frac{1}{2}} }}{2cx^{c-1} + c(c-1)x^{c-2}}. \end{align*}

So, the limit in the numerator is \sqrt{3}, which means we need the denominator to have a finite, non-zero limit (in order for the application of L’Hopital’s rule to be justified and to get a finite, non-zero limit as the problem requests). The only way this can happen is if c = 1 (since if c > 1 then both terms in the denominator are going to 0 as x \to +\infty and if c < 1 then the denominator is going to +\infty as x \to +\infty). Now, plugging in this value of c and evaluating the limit we have

    \begin{align*} \lim_{x \to +\infty} \frac{ \frac{3}{\sqrt{3 + \frac{1}{x^2}}}}{2 \cdot 1 x^{1-1} + 1(1-1)x^{1-2}} &= \lim_{x\to +\infty} \frac{ \frac{3}{\sqrt{3+\frac{1}{x^2}}}}{2} \\[9pt] &= \frac{\sqrt{3}}{2}. \end{align*}

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