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Find constants to make a given limit expression equal to zero

by RoRi

Find values for constants a and b so that

    \[ \lim_{x \to 0} \left( \frac{\sin (3x)}{x^3} + \frac{a}{x^2} + b \right) = 0. \]

We set the limit equal to 0 and solve for the constants, using L’Hopital’s.

    \begin{align*} &&\lim_{x \to 0} \left( \frac{\sin (3x)}{x^3} + \frac{a}{x^2} + b \right) &= 0 \\[9pt] \implies && \lim_{x \to 0} \frac{\sin (3x) + ax + bx^3}{x^3} &= 0 \\[9pt] \implies && \lim_{x \to 0} \frac{3 \cos (3x) + a + 3bx^2}{3x^2} &= 0. \end{align*}

In order for this limit to exist we must have

    \[ \lim_{x \to 0} (3 \cos (3x) + a + 3bx^2) = 0 \]

(otherwise the limit would diverge to +\infty since the denominator is going to 0). Since this is a continuous function (it is the sum of continuous functions) we have

    \[ \lim_{x \to 0} (3 \cos (3x) + a + 3bx^2 = 3 + a \qquad \implies \qquad a = -3. \]

Next, we plug this value of a into the expression above and apply L’Hopital’s rule again and solve for b,

    \begin{align*} &&\lim_{x \to 0} \frac{3 \cos (3x) + a + 3bx^2}{3x^2} &= 0 \\[9pt] \implies && \lim_{x \to 0} \frac{3 \cos (3x) + 3bx^2 - 3}{3x^2} &= 0 \\[9pt] \implies && \lim_{x \to 0} \frac{-9 \sin (3x) + 6bx}{6x} &= 0 \\[9pt] \implies && \lim_{x \to 0} \frac{-3 \sin (3x) + 2bx}{2x} &= 0 \\[9pt] \implies && \lim_{x \to 0} \frac{-9 \cos (3x) + 2b}{2} &= 0 \\[9pt] \implies && \frac{2b-9}{2} &= 0 \\[9pt] \implies && b = \frac{9}{2}. \end{align*}

Therefore, we have

    \[ a = -3, \qquad b = \frac{9}{2}. \]

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