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How can I evaluate this limit? Trying with Taylor doesn't seem to give me the right result, why ?

$$\lim_{x\to 0} \frac{\sqrt{1+4x} -1 -\sin(2x)}{\log(1+x^2)}$$

with Taylor I can approximate $\sin(2x)$ to $ 2x $ and $\log(1+x^2)$ to $ x^2 $. If I plug in these in the limits it does not give me the right limit, what am I doing wrong?

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    $\begingroup$ is it $$\log(1+x^2)$$? $\endgroup$ – Dr. Sonnhard Graubner Oct 9 '17 at 18:48
  • $\begingroup$ Use Taylor at the second oder. The first oder is pointless $\endgroup$ – Guy Fsone Oct 9 '17 at 18:52
  • $\begingroup$ You may have mis-stated the problem: Are you sure you did not mean $\sqrt{1+4x^2}$ rather than $\sqrt{1+4x}$? As it is, the limit diverges. OOps, that is wrog of me but I can't retract the comment. $\endgroup$ – Mark Fischler Oct 9 '17 at 18:52
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$$\log(x^2+1)\sim x^2$$ $$\sin(2x)\sim 2x -\frac{8}{6}x^3$$

$$ \sqrt{1+4x} -1\sim 2x-2x^2 $$ since $$(\sqrt{1+4x} -1)'|_{x=0} = 2~~~and ~~~(\sqrt{1+4x} -1)"|_{x=0} = -4$$

$$\lim_{x\to0} \frac{\sqrt{1+4x} -1 -\sin{2x}}{\log{1+x^2}}\sim \lim_{x\to0}\frac{ -2x^2+\frac{8}{6}x^3}{ x^2} = -2$$

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It is more convenient to use L'Hospital Rule and fact that $\ln(1+x^2)\sim x^2$ as $x\to 0$: $$\lim_{x\to 0}\frac{\sqrt{1+4x}-1-\sin2x}{\ln(1+x^2)}\\=\lim_{x\to 0}\frac{\sqrt{1+4x}-1-\sin2x}{x^2}\\=\lim_{x\to 0}\frac{2(1+4x)^{-1/2}-2\cos2x}{2x}\\=\lim_{x\to 0}\frac{-4(1+4x)^{-3/2}+4\sin2x}{2}=-2$$

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binomial theorem: $(a+b)^n) = a^n + na^{n-1}b + \frac {n(n-1)}{2} a^{n-1}b^2 \cdots$

works for fractional exponents

$(1+4x)^\frac 12 = 1 + \frac 12 (4x) - \frac 18 (4x)^2+ \cdots%$

$\sin x = x - \frac 16 x^2+\cdots\\ \sin 2x = (2x) - \frac16 (2x)^3+ \cdots$

numerator: $1 + 2x - 2x^2 - 1 - 2x + \frac 86 x^3 = -2x^2- \frac 16 x^3 + \cdots$

denominator:

$\ln (1 + x) = x - \frac 12 x^2 + \frac 13 x^3\\ \ln(1+x^2) = x^2 - \frac 12 x^4\cdots$

$-2$

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Taylor series can be used for approximations as well as for evaluation of limits. Why do you want to use Taylor series for approximations when your goal here is to evaluate a limit? For the current question use the following Taylor series expansions $$\log(1+x^{2})=x^{2}+o(x^{2}),\sqrt{1+4x}=1+2x-2x^{2}+o(x^{2}),\sin 2x=2x+o(x^{2})$$


Alternatively, one can just multiply numerator and denominator by $\sqrt{1+4x}+1+\sin 2x$ to get $$\dfrac{4x-2\sin 2x-\sin^{2}2x}{\dfrac{\log(1+x^{2})}{x^{2}}\cdot x^{2}\{\sqrt{1+4x}+1+\sin 2x\}}$$ The first and last factors in denominator tend to $1$ and $2$ respectively and hence the desired limit is equal to the limit of $$\frac{2x-\sin 2x}{x^{2}}-\frac{\sin^{2}2x}{2x^{2}}$$ Putting $2x=t$ we see that $t\to 0$ as $x\to 0$ and the desired limit is equal to the limit of the expression $$4\cdot\frac{t-\sin t} {t^{2}}-2\cdot\frac{\sin^{2}t}{t^{2}}$$ The first expression tends to $0$ and second one tends to $2$ so the desired limit is $-2$.


Note that the limit of $(t-\sin t) /t^{2}$ can be evaluated using the inequality $$\cos t<\frac{\sin t} {t} <1$$ which leads to $$0<\frac{t-\sin t} {t^{2}}<\frac{1-\cos t} {t} $$ for $0<t<\pi/2$. The result now easily follows from Squeeze theorem as $t\to 0^{+}$.

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