Discuss $\lim_{x \to 0} \frac {(\sin{3x})^a \ln{\cos{x}}}{\sqrt{a^2+x}-|a|}$ according to the variation of $a$, without Hopital

Question

The limit in question is $$\lim_{x \to 0} \frac {(\sin{3x})^a \ln{\cos{x}}}{\sqrt{a^2+x}-|a|}$$ I tried to solve it, but the possible cases seems to me too much. I need to consider $a=0, a!=0, a rational and a natural, and still I don't have all the possible cases. Is there a simple solution?

Answer 1

$$\frac{\sin^a3x\log\cos x}{\sqrt{a^2+x}-|a|}=\frac{\sin^a3x\left(\sqrt{a^2+x}+|a|\right)\log\cos x}x=$$

$$=3^ax^{a-1}\frac{\sin^a3x}{(3x)^a}\left(\sqrt{a^2+x}+|a|\right)\log\cos x$$

Now, if we have $\;a=0\;$ then that's simply

$$\frac{\log\cos x}{\sqrt x}=\sqrt x\frac{\log\cos x}x \xrightarrow[x\to0^+]{}\sqrt0\,\left(\log\cos x\right)'_{x=0}=0$$

Of course, if you really want the limit when $\;x\to0\;$ then in the above case it doesn't exist (as real function) since $\;\sqrt x\;$ isn't defined on negative numbers...

Next:

$$a>1\implies3^ax^{a-1}\frac{\sin^a3x}{(3x)^a}\left(\sqrt{a^2+x}+|a|\right)\log\cos x\xrightarrow[x\to0]{}3^a\cdot0\cdot1^a\cdot2|a|=0$$

Try now to complete the other cases...

Answer 2

Following up on DonAntonio's analysis of the expression that we'll rewrite as

$${(\sin3x)^a\ln\cos x\over\sqrt{a^2+x}-|a|}=3^a\left(\sin3x\over3x\right)^a\left(\sqrt{a^2+x}+|a|\right)\left(\log\cos x\over x^2\right)x^{a+1}$$

The reason for the rewrite is that everything now boils down to showing

$$\lim_{x\to0}{\log\cos x\over x^2}=-{1\over2}$$

This is a snap with L'Hopital. Without L'H, it requires some tricks. Here's one way.

To begin, it's convenient to show $-{\log\cos^2x\over x^2}\to1$. Using the definition of the (natural) log as an integral, we have

$$-\log\cos^2x=-\int_1^{\cos^2x}{dt\over t}=-\int_1^{1-\sin^2x}{dt\over t}=\int_0^{\sin^2x}{du\over1-u}$$

Now for $0\le u\le\sin^2x$, we have

$$1\le{1\over1-u}=1+{u\over1-u}\le1+{u\over1-\sin^2x}$$

and thus

$$\sin^2x\le\int_0^{\sin^2x}{du\over1-u}\le\sin^2x+{{1\over2}\sin^4x\over1-\sin^2x}$$

Consequently

$$\left(\sin x\over x\right)^2\le-{\log\cos x\over x^2}\le\left(\sin x\over x\right)^2+{1\over2}\left(\sin x\over x\right)^2{\sin^2x\over1-\sin^2x}$$

Knowing that ${\sin x\over x}\to1$ as $x\to0$ allows the Squeeze Theorem to finish the job.

To summarize the implications for the original question, we have

$$3^a\left(\sin3x\over3x\right)^a\left(\sqrt{a^2+x}+|a|\right)\left(\log\cos x\over x^2\right)\to3^a\cdot1\cdot2|a|\cdot{-1\over2}=-3^a|a|$$

and therefore the OP's expression, which has an additional factor of $x^{a+1}$, tends to $0$ if $a\gt-1$, to $-{1\over3}$ if $a=-1$, and it diverges if $a\lt-1$.