If $\quad \lim _{x \rightarrow 0}\left(\frac{a \sin ^4 \sqrt{x}}{\sqrt{\cos x}-1}\right)^{\frac{\tan \left(\ln ^2(1+\sqrt{2} x)\right)}{\ln \left(1+\sin ^2 x\right)}}=16$, sum of all possible values of $a$ is

On seeing this , the first thing that immediately ran through my mind is limit does not exist. Becasue $\sqrt x$ is given and when x approaches 0 from left side of 0, it take negative values and thus root x cant be defined.

If I'm wrong I don't find any other way to simplify this. Can you help?

  • 3
    $\begingroup$ Your observation about $\sqrt x$ is certainly correct; maybe they mean $\lim_{x\to0^+}$. $\endgroup$ Jun 6 at 4:12
  • $\begingroup$ Where did this question come from? Additional context will help in knowing (a) if it's meant to have a solution in the first place, (b) what kind of methods are expected to be used to solve it. (Assuming that we get around the problems like the one you already identified, that is.) $\endgroup$
    – ConMan
    Jun 6 at 4:42
  • $\begingroup$ @ConMan This question was created by my professor. He usually creates questions for various top level examinations in Asia $\endgroup$ Jun 6 at 14:18

1 Answer 1


Actually in this case unlike your other question the limit can be considered from both directions without confusion. Observe that

$$\sin^4\sqrt{x} = (1-\cos^2\sqrt{x})^2$$

where we made the change to $\cos$ to note more easily that in the case of $x<0$, $\cos^2\sqrt{-x} = \cosh^2\sqrt{x}$ without any ambiguity or choice of branch in the complex plane. You can algebraically prove this for yourself. This expression will always evaluate to a single (not multivalued) real number.

Having said that, consider rewriting the limmand as

$$\left(a\cdot \frac{\sin^4\sqrt{x}}{x^2}\cdot\frac{x^2}{\cos x - 1}\cdot(\sqrt{\cos x} + 1)\right)^{\frac{\tan\left(\log^2(1+\sqrt{2}x)\right)}{\log^2(1+\sqrt{2}x)}\cdot\frac{\log^2(1+\sqrt{2}x)}{2x^2}\cdot\frac{x^2}{\sin^2 x}\cdot\frac{\sin^2 x}{\log(1+\sin^2 x)}\cdot 2}$$

where we have multiplied by a series of terms of the form $\frac{f(x)}{f(x)}$ leaving the original value of the function unaltered. If taken, the limit in the exponent evaluates to

$$\lim_{x\to0}\frac{\tan\left(\log^2(1+\sqrt{2}x)\right)}{\log^2(1+\sqrt{2}x)}\cdot\frac{\log^2(1+\sqrt{2}x)}{2x^2}\cdot\frac{x^2}{\sin^2 x}\cdot\frac{\sin^2 x}{\log(1+\sin^2 x)}\cdot 2 = 1\cdot 1 \cdot 1 \cdot 1 \cdot 2 = 2$$

by using the known limits of $\frac{\tan u}{u}$, $\frac{\log(1+u)}{u}$, and $\frac{\sin u}{u}$ as $u\to 0$ . This would mean that the limit for the base must approach either $\pm 4$ (but we will get to the feasibility of both options later). Now ready to tackle the limit of the base, we get

$$\lim_{x\to0}a\cdot \frac{\sin^4\sqrt{x}}{x^2}\cdot\frac{x^2}{\cos x - 1}\cdot(\sqrt{\cos x} + 1) = a \cdot 1 \cdot -2 \cdot 2 = -4a$$

using the known limits $\frac{\sin u}{u}$ and $\frac{1-\cos u}{u^2}$.

From here it seems like a simple matter that our choices for $a$ from what we determined earlier are $\pm 1$. Unlike the last time, attempting to make the base negative will make the full exponential evaluation multivalued since the exponent is an arbitrary real function instead of being restricted to the integers. This means we have to ensure the base is positive in a neighborhood of $0$, and there is only one possible value, $\boxed{a=-1}$, that meets this criteria.

  • $\begingroup$ Thank you Ninad. I really appreciate your efforts to solve me ques, Just a doubt. I just entered high school. So I don't have any clue on what cosh is? Could you tell me what that cosh is? $\endgroup$ Jun 6 at 14:21
  • $\begingroup$ @ElizabethHuffman cosh(x) = (exp(x)+exp(-x))/2 = cos(ix) $\endgroup$ Jun 6 at 14:29
  • $\begingroup$ @ElizabethHuffman: $\cosh$ is the hyperbolic cosine function. $\endgroup$ Jun 6 at 14:29
  • $\begingroup$ @MichaelSeifert Thank you Michael. Like I havent learnt hyperbolic function. But do you think its knowledge is necessary to solve this question? $\endgroup$ Jun 6 at 14:30
  • $\begingroup$ @JuliaHayward Thank you so much Julia. I appreciate your time. $\endgroup$ Jun 6 at 14:30

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