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What is the limit of this function as $x \rightarrow 0$ ?

$$\lim_{x\rightarrow 0} \frac{36^x-9^x-4^x+1}{\sqrt 2 -\sqrt{1+\cos(x)}}$$

I have tried simplifying the expression in different ways. For instance, taking the natural logarithm or multiplying the numerator and denominator by the conjugate of the radical term. But they are not helping me in any way.

According to Wolframalpha the answer of this question is:

$4 \sqrt 2 \log(4) \log(9)$.

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    $\begingroup$ Rewrite $n^x$ as $\exp(x\log n)$ and Taylor expand? $\endgroup$ – user88595 May 30 '15 at 15:58
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    $\begingroup$ Instead of Taylor expanding, you could also use L'Hopital's rule, not such which is less messy. $\endgroup$ – user88595 May 30 '15 at 15:59
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HINT:

$$=\dfrac{9^x-1}x\cdot\dfrac{4^x-1}x\cdot\dfrac{\sqrt2+\sqrt{1+\cos x}}{\dfrac{1-\cos x}{x^2}}$$

Now $\lim_{h\to0}\dfrac{e^h-1}h=1$

For $a>0,a^h=e^{h\ln a}, \lim_{h\to0}\dfrac{a^h-1}h=\cdots=\ln a$

Again, $\lim_{x\to0}\dfrac{1-\cos x}{x^2}=\left(\lim_{x\to0}\dfrac{\sin x}x\right)^2\cdot\dfrac1{\lim_{x\to0}(1+\cos x)}=\cdots$

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With Taylor expansions: $$\begin{align} 36^x - 9^x - 4^x +1 &= e^{x\ln 36} - e^{x\ln 9} - e^{x\ln 4} + 1 \\ &= 1+x\ln 36 + \frac{x^2}{2}\ln^2 36 - (1+x\ln 9 + \frac{x^2}{2}\ln^2 9) -(1+x\ln 4 + \frac{x^2}{2}\ln^2 4)+1 + o(x^2)\\ &= \frac{x^2}{2}\ln^2 36 - \frac{x^2}{2}\ln^2 9 - \frac{x^2}{2}\ln^2 4 + o(x^2) = 2x^2\cdot(\ln^2 6-\ln^2 3 - \ln^2 2) + o(x^2) \end{align}$$ where we went to order $x^2$ (and could simplify the $x$'s) as $\ln 36 = \ln 9 + \ln 4$.

Now, for the denominator:

$$\begin{align} \sqrt{2} - \sqrt{1+\cos x} &= \sqrt{2} - \sqrt{1+1-\frac{x^2}{2}+o(x^2)} = \sqrt{2} \cdot \left(1- \sqrt{1-\frac{x^2}{4}+o(x^2)}\right) \\ &= \sqrt{2} \cdot \left(1- \left(1-\frac{x^2}{8}+o(x^2)\right)\right) = \sqrt{2} \cdot \left(\frac{x^2}{8}+o(x^2)\right) \end{align}$$

so that taking the ratio to put everything together, you obtain $$ \frac{2x^2\cdot(\ln^2 6-\ln^2 3 - \ln^2 2) + o(x^2)}{\sqrt{2} \cdot \left(\frac{x^2}{8}+o(x^2)\right)} = \frac{16\cdot(\ln^2 6-\ln^2 3 - \ln^2 2) + o(1)}{\sqrt{2}+o(1)} \xrightarrow[x\to0]{} \frac{16\cdot(\ln^2 6-\ln^2 3 - \ln^2 2)}{\sqrt{2}} $$

Now, this last quantity can be shown to be equal to $16\sqrt{2}\ln 2\ln 3$ using properties of the logarithm, which is the result Wolfram gave you. Indeed, $$\begin{align} (\ln ^2 6 - \ln^2 3) - \ln^2 2 &= (\ln 6 + \ln 3) (\ln 6 - \ln 3) - \ln^2 2\\ &= (\ln 18) (\ln 2) - \ln^2 2 = (\ln 18 - \ln 2) \cdot \ln 2 = \ln 9 \cdot \ln 2 \\&= 2\cdot \ln 3 \cdot \ln 2. \end{align}$$

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  • $\begingroup$ This is not the most elegant way to do it (see lab bhattacharjee's answer for a more succinct proof), but it has the advantage of being systematic and does generalize (a Taylor expansion approach is not necessarily beautiful nor enjoyable, but does the job most of the time in my experience). Also, compared to a third possibility, it does not use L'Hopital, which I loathe. $\endgroup$ – Clement C. May 30 '15 at 16:20
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This is a mild variant of lab bhattacharjee's answer. Using $\cos 2t=2\cos^2 t-1=1-2\sin^2 t$, we can rewrite $\sqrt{2}-\sqrt{1+\cos x}$ as $\sqrt{2}(1-\cos(x/2))$ and then as $2\sqrt{2}\sin^2(x/4)$. So our limit is $$\lim_{x\to 0}\frac{9^x-1}{x}\cdot\frac{4^x-1}{x}\cdot\frac{16}{2\sqrt{2}}\cdot\left(\frac{x/4}{\sin(x/4)}\right)^2.$$

We recognize $\lim_{x\to 0} \frac{9^x-1}{x}$ as the derivative of $9^x$ at $x=0$. Similarly, $\lim_{x\to 0}\frac{4^x-1}{x}$ is the derivative of $4^x$ at $x=0$. And finally, $\lim_{x\to 0}\frac{x/4}{\sin(x/4)}=1$.

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