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I have to calculate limit $$\lim_{x\to 1 } \left(\frac{2^x x + 1}{3^x x}\right)^{\tan(\frac{\pi x}{2})}.$$ I know $\tan(\frac{\pi x}{2})$ is undefined in $x = 1$, but can I just put $x = 1$ into $\frac{x\cdot 2^x + 1}{x\cdot3^x}$ and get $$\lim_{x\to 1 } (1)^{\tan(\frac{\pi x}{2})} = 1.$$ Is the answer $1$ correct? It's forbidden to use L'Hôpital's rule.

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    $\begingroup$ No, you can not $\endgroup$ – Val Nov 18 '15 at 7:59
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    $\begingroup$ Limits are not about evaluating functions in specific points. They are about seeing what the function does as we get closer and closer to that point. You can directly substitute only when the function is continuous in a neighborhood of that point. $\endgroup$ – rubik Nov 18 '15 at 8:03
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For $x$ near $0$, $a^x=1+x\log(a)+O\left(x^2\right)$.

Furthermore, if $\lim\limits_{n\to\infty}\left|b_n\right|=\infty$ and $c=\lim\limits_{n\to\infty}a_nb_n$, then $\lim\limits_{n\to\infty}\left(1+a_n\right)^{b_n}=e^c$.

Therefore, $$ \begin{align} \lim_{x\to1}\left(\frac{2^xx+1}{3^xx}\right)^{\tan\left(\frac{\pi x}2\right)} &=\lim_{x\to0}\left(\frac{2^{x+1}(x+1)+1}{3^{x+1}(x+1)}\right)^{\tan\left(\frac\pi2(x+1)\right)}\\ &=\lim_{x\to0}\left(\frac{2(1+x)\left(1+x\log(2)+O\left(x^2\right)\right)+1}{3(1+x)\left(1+x\log(3)+O\left(x^2\right)\right)}\right)^{-1/\tan\left(\frac\pi2x\right)}\\ &=\lim_{x\to0}\left(\frac{3+x(2+2\log(2))+O\left(x^2\right)}{3+x(3+3\log(3))+O\left(x^2\right)}\right)^{-1/\tan\left(\frac\pi2x\right)}\\ &=\lim_{x\to0}\left(1-\frac x3\left(1+\log\left(\frac{27}4\right)\right)+O\left(x^2\right)\right)^{-1/\tan\left(\frac\pi2x\right)}\\[9pt] &=e^{\frac2{3\pi}\left(1+\log\left(\frac{27}4\right)\right)} \end{align} $$

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Two hints: first take $\ln$, second, do a change of variable $h=x-1$. $$ \ln L = \lim_{x\to 1}\tan\left(\frac{\pi x}2\right)\ln\left( \frac{x 2^x + 1}{x 3^x}\right) = \lim_{h\to 0}\tan\left(\frac{\pi(h+1)}2\right)\ln\left( \frac{(h+1)2^{h+1} + 1}{(h+1)3^{h+1}}\right)=\cdots $$ Apply the trigonometric formula for the tangent of a sum...

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  • $\begingroup$ I don't see where to do variable change. How limit changes when we take log? $\endgroup$ – Desh Nov 18 '15 at 8:15
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    $\begingroup$ @Denis, $\ln\lim=\lim\ln$ because $\ln$ is continuous and if $h=x−1$, then $x=h+1$. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 18 '15 at 8:16
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Firstly, your reasoning leads to $\lim_{x\to 0} \frac{x}{x} = \lim_{x\to 0} \frac{0}{x} = 0$ 'because you can just put $x=0$ into $x$'. But of course it's wrong; you cannot replace part of an expression with something else that isn't equal. Indeed the limit is more-or-less defined as the value (if it exists) that you would eventually approach as $x$ gets closer but never reaches $0$.

With that understood, you always want to express $a^b = \exp(b \cdot \ln(a))$ so that you can use Taylor expansion. And usually it's easier to understand the behaviour and expand around $0$, so we should compute $\lim_{d\to 0} \left(\dfrac{2^{1+d} (1+d) + 1}{3^{1+d} (1+d)}\right)^{\tan(\frac{\pi (1+d)}{2})}$ instead. The method I will use is asymptotic expansion using Landau's Little-O-notation, which I would encourage you to learn, as it applies to any limit problem in general. $\def\wi{\subseteq}$

Basic asymptotic expansions

$\exp(x) \in 1+x+o(x)$ as $x \to 0$.

$\ln(1+x) \in x+o(x)$ as $x \to 0$.

$a^x = \exp(x\ln(a)) \in 1+x\ln(a)+o(x)$ as $x \to 0$, for any $a > 0$.

You can of course use more terms from the respective Taylor expansions if the first-order terms cancel and are not enough.

Solution

As $d \to 0$:

  $\left(\dfrac{2^{1+d} (1+d) + 1}{3^{1+d} (1+d)}\right)^{\tan(\frac{\pi (1+d)}{2})}$

  $= \exp\left( \tan(\frac{\pi (1+d)}{2}) \ln\!\left(\dfrac{2^d 2(1+d) + 1}{3^d 3(1+d)}\right) \right)$

  $\in \exp\left( \dfrac{1}{\tan(-\frac{\pi}{2}d)} \ln\!\left(\dfrac{(1+d\ln(2)+o(d)) 2(1+d) + 1}{(1+d\ln(3)+o(d)) 3(1+d)}\right) \right)$

  $\wi \exp\left( \dfrac{1}{-\frac{\pi}{2}d+o(d)} \ln\!\left(\dfrac{3+(2\ln(2)+2)d+o(d)}{3+(3\ln(3)+3)d+o(d)}\right) \right)$

  $\wi \exp\left( \dfrac{1}{-\frac{\pi}{2}d} (1+o(1)) \ln\!\Big(1-\frac{3\ln(3)-2\ln(2)+1}{3}d+o(d)\Big) \right)$

  $\wi \exp\left( \dfrac{1}{-\frac{\pi}{2}d} (1+o(1)) \Big(-\frac{3\ln(3)-2\ln(2)+1}{3}d+o(d)\Big) \right)$

  $\wi \exp\left( \dfrac{2}{\pi} (1+o(1)) \Big(\frac{3\ln(3)-2\ln(2)+1}{3}+o(1)\Big) \right)$

  $\wi \exp\left( \dfrac{2(3\ln(3)-2\ln(2)+1)}{3\pi} + o(1) \right)$

  $\wi \exp\left( \dfrac{2(3\ln(3)-2\ln(2)+1)}{3\pi} \right) \exp(o(1))$

  $\to \exp\left( \dfrac{2(3\ln(3)-2\ln(2)+1)}{3\pi} \right)$.

Comments

This kind of technique is what computer algebra systems do as well, such as Wolfram Alpha.

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Let $$y = \left( \frac{x 2^x + 1}{x 3^x}\right)^{\tan\left(\frac{\pi x}{2}\right)}$$ Then $$\ln(y) = \tan\left(\frac{\pi x}{2}\right)\ln\left( \frac{x 2^x + 1}{x 3^x}\right) = \frac{\ln\left( \frac{x 2^x + 1}{x 3^x}\right)}{\frac{1}{\tan\left(\frac{\pi x}{2}\right)}}$$ Apply L'Hôpital's rule to obtain your answer, it is still a lot of work. Maybe there is a faster way to go about this problem.

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  • $\begingroup$ It's forbidden to use L'Hôpital's rule $\endgroup$ – Desh Nov 18 '15 at 8:12
  • $\begingroup$ Would be good to mention that at the start then... $\endgroup$ – Ian Miller Nov 18 '15 at 8:14
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    $\begingroup$ In your last expression, write the top as $\ln f(x),$ the bottom as $g(x)=\cot (\pi x/2).$ If we divide top and bottom by $x-1,$ then the expression $\to (\ln f)'(1)/g'(1)$ by the definition of the derivative. No L'Hopital involved at all. $\endgroup$ – zhw. Nov 18 '15 at 8:40

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