How to evaluate $\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+sinx}-3}$?

Question

$$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$ I have tried with Taylor: $$\lim _{x\to \:0^+}\frac{\left(1+x+\frac{x^2}{2}+1+2x+2x^2\right)^2-4}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}=\lim _{x\to \:0^+}\frac{\frac{25x^4}{4}+15x^3+19x^2+12x}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}$$ But now I dont know how to move forward.

I tried with Hopital and it was found to be so: $$\lim _{x\to \:0+}\left(\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin \left(x\right)}-3}\right)=\lim _{x\to \:0+}\left(\frac{2e^{2x}\left(e^x+1\right)\left(2e^x+1\right)}{\frac{\cos \left(x\right)}{2\sqrt{\sin \left(x\right)+9}}}\right)$$ $$=\lim _{x\to \:0+}\left(\frac{4e^{2x}\left(e^x+1\right)\left(2e^x+1\right)\sqrt{\sin \left(x\right)+9}}{\cos \left(x\right)}\right)=\frac{4e^{2\cdot \:0}\left(e^0+1\right)\left(2e^0+1\right)\sqrt{\sin \left(0\right)+9}}{\cos \left(0\right)}=\color{red}{72}$$ But I wanted to know if you could get the same result even with Taylor? Thanks

Answer 1

$$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$

$$=\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sin x}\cdot\lim _{x\to \:0^+}(\sqrt{9+\sin x}+3)$$

$$=\left(\lim _{x\to0^+}\frac{e^{2x}-1}x+\lim _{x\to0^+}\frac{e^{4x}-1}x+2\lim _{x\to0^+}\cdot\frac{e^{3x}-1}x\right)\cdot\dfrac1{\lim _{x\to0^+}\dfrac{\sin x}x}\cdot\lim _{x\to0^+}(\sqrt{9+\sin x}+3)$$

Now use $\lim_{h\to0}\dfrac{e^{mh}-1}h=m$ to get $$(2+4+2\cdot3)\cdot(\sqrt9+3)=72$$

Answer 2

To simplify, let us first factor $(e^x+e^{2x})^2-4=(e^x+e^{2x}-2)(e^x+e^{2x}+2)$ and get rid of the square root with $(\sqrt{9+\sin x}-3)(\sqrt{9+\sin x}+3)=\sin x$.

Then

$$\frac{e^x+e^{2x}-2}{\sin x}=\frac{1+x+\frac12x^2\cdots+1+2x+2x^2\cdots-2}{x-\frac16x^3\cdots}=\frac{3x+\frac52x^2\cdots}{x-\frac16x^3\cdots}$$tends to $3$.

The final answer is

$$(1+1+2)(\sqrt9+3)\,3=72.$$

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The fully painful way, with development up to second order yields

$$\frac{(1+x+\frac12x^2\cdots+1+2x+2x^2\cdots)^2-4}{\sqrt{9+x-\frac16x^3\cdots}-3}\\ =\frac{4+12x+19x^2\cdots-4}{\sqrt9+\frac1{2\sqrt9}(x-\frac16x^3\cdots)-\frac1{2\cdot2\cdot9\sqrt9\cdot2}(x-\frac16x^3\cdots)^2\cdots-3}=\frac{12x+19x^2\cdots}{\frac16x-\frac1{216}x^2\cdots}$$ which tends to $12\cdot6=72$.

[We used $\sqrt{a+x}=\sqrt a+\frac1{2\sqrt a}x-\frac1{2\cdot2\cdot a\sqrt a\cdot2}x^2\cdots$]