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I have the following task:

For what values of $a$ and $b$ is the following equation true? $$\lim\limits_{x \to 0} \left(\frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}\right) = 0 $$

I want to know the steps I should follow in order to find the solution.

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    $\begingroup$ Your question title is not very informative. $\endgroup$
    – N.S.JOHN
    Feb 23 '16 at 11:21
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We have \begin{align} L &= \lim_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2} \\ &= \lim_{x \to 0} \frac{\sin(2x) + a x^3 + b x }{x^3}\\ \end{align} which is a limit of type $0 / 0$, so we try L'Hôpital's rule:

\begin{align} L &= \lim_{x \to 0} \frac{2\cos(2x) + 3 a x^2 + b}{3x^2} \end{align}

The denominator again vanishes for $x \to 0$, the nominator goes to $2 + b$. So if $b \ne -2$, the nominator does not vanish and we have $$ \DeclareMathOperator{sgn}{sgn} L = \sgn(2 + b) \, \infty $$ For $b = -2$ we again have a limit of type $0 / 0$ and apply the rule again: \begin{align} L &= \lim_{x\to 0} \frac{-4 \sin(2x) + 6ax}{6x} \end{align} The nominator and denominator vanish and we apply the rule once again:

\begin{align} L &= \lim_{x\to 0} \frac{-8 \cos(2x) + 6a}{6} = \frac{6a-8}{6} = a - \frac{4}{3} \end{align}

This gives the answer that the equation is true, $L$ vanishes, if $a = 4/3$ and $b = -2$.

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If you use L'Hôpital's rule on $\frac{\sin(2x)}{x^3}$ and $\frac{b}{x^2}$ you will see that neither has a limit as $x$ approaches zero. Therefore you cannot split your full limit into separate limits to get your answer.

Put the expression inside the limit into a single fraction, with denominator $x^3$. Ensure that the expression has the form $\frac 00$. Then use L'Hôpital's rule to find its limit. If that does not have a clear limit, again make sure the expression has a proper form for L'Hôpital's rule, which may put limits on $a$ and/or $b$, then use L'Hôpital's rule again.

Keep doing this until you get a clear limit involving $a$ and/or $b$. Set that limit to zero and solve for $a$ and/or $b$.

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$\lim_{x\to0}\frac{\sin 2x}{2x}=1$. The rest of the terms $\frac{2}{x*x}+a+\frac{b}{x*x}=\frac{2+b}{x*x}+a=0$

So $b=-2$, $a=0$

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