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We need to solve limit: $$\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4\cos^2{x}-1}$$ without using derivation (L'Hospital's rule). With the substitution $t = 4\cos^2{x}-1$, I got it to $$\lim_{t \to 0} \frac{\ln{(\sqrt{9-3t}-2)}}{t}$$ but can't progress any further without going in circles.

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    $\begingroup$ For the last limit, shouldn't it be $t$ which tends to $0$? $\endgroup$
    – Bernard
    Sep 15, 2021 at 21:25
  • $\begingroup$ It is corrected. $\endgroup$
    – Comizard
    Sep 15, 2021 at 21:59
  • $\begingroup$ On the surface, unless I misunderstood the constraints your teacher is imposing, I don't really see a way around an epsilon delta proof unless you already have some list of standard limits you are allowed to use. $\endgroup$ Sep 16, 2021 at 8:35

4 Answers 4

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If $f(t)=\ln\left(\sqrt{9-3t}-2\right)$, then $f(0)=0$ and\begin{align}\lim_{t\to0}\frac{\ln\left(\sqrt{9-3t}-2\right)}t&=\lim_{t\to0}\frac{f(t)-f(0)}t\\&=f'(0)\\&=-\frac12.\end{align}

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  • $\begingroup$ Very nice and instantaneous $\endgroup$
    – Sebastiano
    Sep 15, 2021 at 21:30
  • $\begingroup$ This is a nice solution and I like it, but i don't know if I am allowed to do any kind of diferentiation and here I still need to calculate f'(0). Really not sure how this can be done by only using tabular limits but formulation demands it. Thanks anyway! $\endgroup$
    – Comizard
    Sep 15, 2021 at 21:58
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Do another step and substitute $$ \sqrt{9-3t}-2=1+u $$ whereby $$ 9-3t=u^2+6u+9 $$ and $t=-(u^2+6u)/3$, so your limit becomes $$ \lim_{u\to0}\frac{\ln(1+u)}{u}\frac{3}{-u-6} $$

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  • $\begingroup$ Nice and simple. Thanks! $\endgroup$
    – Comizard
    Sep 16, 2021 at 10:59
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If you are not to use differentiation in any form (l'Hôpital's Rule, direct derivative, Taylor series, etc.) you need to know the "rule"

$\lim_{u\to0}\dfrac{\ln(1+u)}{u}=1$

Here, render $u=\sqrt{9-3t}-3$ and then your limit becomes

$\lim_{t\to0}\left(\dfrac{\ln(\sqrt{9-3t}-2)}{\sqrt{9-3t}-3}\right)\left(\dfrac{\sqrt{9-3t}-3}{t}\right)$

The first factor in large parentheses has the limit $1$ by our logarithm rule. The second factor is handled by multiplying the numerator and denominator by the conjugate factor $\sqrt{9-3t}+3$ and then treating the numerator as a difference of squares:

$(\sqrt{9-3t}-3)(\sqrt{9-3t}+3)=(9-3t)-9=-3t$

which can be used to cancel the factor of $t$ in the denominator and enable direct evaluation at $t=0$.

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Here's another way. First use $\cos^2{x}=1-\sin^2{x}$ and

let $y=\ln{(2\sqrt{3}\sin{x}-2})$, $\frac{e^{y}+2}{2\sqrt{3}}=\sin{x}$, $y\to0$ as $x\to\frac{\pi}{3}$ $$\begin{align} \lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4\cos^2{x}-1} &=\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4(1-\sin^2{x})-1}\\ &=\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{3-4\sin^2{x}}\\ &=\lim_{y \to 0} \frac{y}{3-4 \left(\frac{e^{y}+2}{2\sqrt{3}}\right)^2}\\ %&=\lim_{y \to 0} \frac{y}{3-4 \left(\frac{e^{2y}+4e^{y}+4}{4\times3}\right)}\\ &=\lim_{y \to 0} \frac{3y}{5-\left(e^{2y}+4e^{y}\right)}\\ &=\lim_{y \to 0} \frac{3y}{5-\left(\displaystyle\sum_{n=0}^{\infty}\frac{(2y)^n}{n!}+4\displaystyle\sum_{n=0}^{\infty}\frac{y^n}{n!}\right)}\\ &=\lim_{y \to 0} \frac{3y}{5-\displaystyle\sum_{n=0}^{\infty}\left(\frac{(2y)^n+4y^n}{n!}\right)}\\ &=\lim_{y \to 0} \frac{3y}{5-\left(\frac{(2y)^0+4y^0}{0!}+\frac{(2y)^1+4y^1}{1!}+\displaystyle\sum_{n=2}^{\infty}\frac{(2y)^n+4y^n}{n!}\right)}\\ &=\lim_{y \to 0} \frac{3y}{5-\left(\frac{1+4}{1}+\frac{2y+4y}{1}+\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}\right)}\\ &=\lim_{y \to 0} \frac{3y}{5-\left(5+6y+\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}\right)}\\ &=\lim_{y \to 0} \frac{3y}{-6y-\displaystyle\sum_{m=0}^{\infty}\left(\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}\right)}\\ &=\lim_{y \to 0} \frac{1}{-2-\frac{1}{3y}\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}}\\ &=\lim_{y \to 0} \frac{1}{-2-\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+1}+4y^{m+1}}{3(m+2)!}}\\ &=\frac{1}{-2}\\ \end{align}$$

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