1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ How about first removing square root in the denominator? $\endgroup$ – Hwang Jan 27 '16 at 9:24
3
$\begingroup$

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

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

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

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

$\endgroup$
  • $\begingroup$ Why becomes so $$\left(\lim _{x\to0^+}\frac{e^{2x}-1}x+\lim _{x\to0^+}\frac{e^{4x}-1}x+2\lim _{x\to0^+}\cdot\frac{e^{3x}-1}x\right)$$ $\endgroup$ – Amarildo Jan 27 '16 at 9:32
  • $\begingroup$ @AmarildoAliaj, What is $$(e^x+e^{2x})^2?$$ $\endgroup$ – lab bhattacharjee Jan 27 '16 at 9:33
  • $\begingroup$ $$\left(e^x+e^{2x}\right)\left(e^x+e^{2x}\right)$$ $\endgroup$ – Amarildo Jan 27 '16 at 9:37
  • $\begingroup$ @AmarildoAliaj, Why don't you expand using $(a+b)^2$ $\endgroup$ – lab bhattacharjee Jan 27 '16 at 9:39
  • $\begingroup$ I dont understand. Can you show me how? Thanks $\endgroup$ – Amarildo Jan 27 '16 at 9:41
0
$\begingroup$

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

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


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.