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What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$

We have to make this function defined, by change the cosine into sine or tan. So,

$$\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}\right) = \lim_{t \to 0} \left(\frac{a}{t^2} - \frac{6}{t^2}\right) = -18$$

But I am confused what should I do next.

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  • $\begingroup$ Your identity between limits is wrong, because if you set $a=6$, it turns to $0=−18$. $\endgroup$ – Yves Daoust Feb 10 at 10:43
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I think the best way to solve your problem is:

Your limit is equal to $\lim_{t \to 0} \frac{at-2\tan (3t)}{t^3}$ which is indeterminate. By applying L'Hôpital rule we have:

$$\lim_{t \to 0} \frac{a-6\sec^2(3t)}{3t^2}$$

Well, this limit is always $\infty$, unless maybe when $a=6$ (for what it is again $\frac{0}{0}$). Applying L'Hôpital again and we have:

$$\lim_{t \to 0} \frac{6-6\sec^2(3t)}{3t^2}=-18\lim_{t \to 0} \frac{1-\sec^2(3t)}{3\times 3\times t^2}=-18\lim_{t \to 0} \frac{\tan^2(3t)}{(3t)^2}$$

Can you figure the remaining out? Is quite simple...

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  • $\begingroup$ Why you said apply LHopital again? You just applied LHopital once? $\endgroup$ – Lifeforbetter Feb 10 at 9:50
  • $\begingroup$ You have to apply it again on $\lim_{t \to 0} \frac{\tan(3t)}{(3t)}$ at the very end. Of course this is a special case, but you don't have to know that. $\endgroup$ – Pspl Feb 10 at 9:52
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What is $a$

It's a free parameter, i.e. the limes is a function of $a$: $$ f(a) = \lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) $$

You'll have to take care for the special case $a=6$.

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  • $\begingroup$ Ok, I relaxed my statement. I assumed her calculation was right... $\endgroup$ – emacs drives me nuts Feb 10 at 9:39
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There's an error in your computation, and you should find first a Laurent expansion: $$\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}=\frac 1{t^2}\biggl(at-2\frac{\tan 3t}t\biggr)$$ Now the Taylor expansion of $\tan x$ at order $3$ is $\;\tan x=x+\dfrac{x^3}{3}+o\bigl(x^3\bigr)$, so $$ \frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}=\frac 1{t^2}\biggl(a-\frac 2t\Bigl(3t+9t^3+ o\bigl(t^3\bigr)\Bigr)\biggr) =\frac{a-6}{t^2} -18+o(1). $$ We conclude there's a limit at $0$ when $a=6$, and in this case the limit is indeed $-18$.

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The cosine tends to $1$ and the sine can be replaced by its argument so that

$$\frac a{t^2}-\frac{\sin t}{t^3\cos^23t}=\frac{a-6}{t^2}+\cdots$$ where the dots denote terms of a higher degree.

So if the question is asking the value of $a$, a necessary condition is

$$a=6$$

and it is certainly fulfilled, as the limit is known to exist.


The question does not seem to check that the value of the limit is indeed $-18$, so you can stop there.

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Just use Taylor expansion and simplify first $\cos^2 3t = \frac{1+\cos 6t}{2}$.

Hence,

$$\frac{at\cos^2 3t - \sin 6t}{t^3\cos^2 3t}= \frac{at(1+\cos 6t) - 2\sin 6t}{t^3(1+\cos 6t)}$$

$$= \frac{at + at(1- 18t^2+o(t^3)) - 2(6t-36t^3+o(t^4))}{t^3(1+\cos 6t)}$$

$$= \frac{1}{(1+\cos 6t)}\left(\frac{2a-12}{t^2} -18a+72 + o(t)\right)$$

So, the limit exists only if $\boxed{a=6}$ and then the limit is equal to $\boxed{\frac{-36}{2}}=-18$.

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  • $\begingroup$ Ooops, forgive me. For some reason I commented to your post by mistake. Fixing... $\endgroup$ – Yves Daoust Feb 10 at 10:43
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Your condition implies that $$\lim_{t\to 0}\frac{at\cos^23t-\sin 6t}{t^3}=-18$$ Adding and subtracting $at$ in numerator we get $$\lim_{t\to 0} \frac{at-\sin 6t}{t^3}-9a\cdot\frac{\sin^23t}{(3t)^2}=-18$$ or $$\lim_{t\to 0}\frac {a-6}{t^2}+\frac{6t-\sin 6t}{t^3}=9(a-2)$$ Using substitution $x=6t$ the second term on left side can be written as $$6^3\lim_{x\to 0}\frac {x-\sin x} {x^3}=36$$ (via L'Hospital's Rule or Taylor series) and therefore we have $$\lim_{t\to 0}\frac {a-6}{t^2}=9(a-6)$$ which implies $a=6$ and the above equation also holds true for this value of $a$.

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