Find limit without L' Hopital's rule?

Question

I will be thankful if somebody help me to find limit without L Hospitals rule for

$$\lim_{x \to \pi}\left[\tan\left(x \over 4\right)\right] ^{1/\left[2\left(x- \pi\right)\right]} $$

Answer 1

Following Michael's suggestion, if L'Hospital's rule is not allowed, perhaps the derivative is? Because then we can rewrite his expression. First sub $2x=t$ so that the limit can be taken for $t$ to $2\pi$ on the expression $\frac{ln(tan0.125t)}{t-2\pi}$. Now in that numerator we can "add" the term $ln(tan\frac{\pi}{4})$ which is really zero (where did I get this from?) and thus we can use the definition of the derivative, in this case $ln(tan\frac{t}{8})$. I want you now to do that work and then plug in for $t=2\pi$. Don't forget to raise it back to the power $e$ to get the actual limit ($e^{\frac{1}{4}}$)

Answer 2

Note that, assuming the latter limit exists, $$L=\lim_{x\to\pi}\left(\tan\frac{x}4\right)^\tfrac{1}{2(x-\pi)}=\lim_{x\to\pi}\left(\tan\frac{x}4\right)^{\tfrac1{\tan\frac x4-1}\tfrac{\tan\frac x4-1}{2(x-\pi)}}=e^{\lim_\limits{x\to\pi}\tfrac{\tan\frac x4-1}{2(x-\pi)}}$$ holds since the function $(x,y)\mapsto x^y$ is continuous ($x>0,y\in\mathbb R$).

Using the substitution $y=x-\pi$ and the addition formula for the tangent function, we now have $$\lim_{x\to\pi}\frac{\tan\frac x4-1}{2(x-\pi)}=\lim_{y\to0}\frac{\tan\frac{y+\pi}4-1}{2y}=\lim_{y\to0}\frac{\frac{\tan\frac y4+1}{1-\tan\frac y4}-1}{2y}=\lim_{y\to0}\frac{\tan\frac y4}{y}\lim_{y\to0}\frac1{1-\tan\frac y4}=\frac14,$$ so the final result is $L=e^{\frac14}$.

Answer 3

Hint: $$\ln\biggl(\tan\left(\frac x 4\right)\big(x- \pi)gr)^\tfrac1{2(x- \pi)}=\frac{\ln\Bigl(\tan\left(\dfrac x 4\right)\Bigr)}{2(x- \pi)}=\frac12\frac{\ln\Bigl(\tan\left(\dfrac x 4\right)\Bigr)-\ln\Bigl(\tan\left(\dfrac \pi 4\right)\Bigr)}{x- \pi}$$ is half the rate of variation of the function $$f(x)=\ln\Bigl(\tan\Bigl(\dfrac x4\Bigr)\Bigr)$$ (starting from $\frac\pi4$). Hence the limit is $$\biggl(\ln\Bigl(\tan\Bigl(\dfrac x4\Bigr)\Bigr)\biggr)'_{x=\tfrac\pi4}.$$

Answer 4

In this answer, it is shown that if $\lim_{x\to0}xy(x)=a$, then $$ \lim_{x\to0}(1+x)^y=e^a\tag{1} $$ Therefore, since $$ \begin{align} \tan\left(\frac{x+\pi}4\right) &=\frac{1+\tan\left(\frac x4\right)}{1-\tan\left(\frac x4\right)}\\ &=1+\frac{2\tan\left(\frac x4\right)}{1-\tan\left(\frac x4\right)}\tag{2} \end{align} $$ and $$ \lim_{x\to0}\frac{2\tan\left(\frac x4\right)}{1-\tan\left(\frac x4\right)}\frac1{2x}=\frac14\tag{3} $$ we get $$ \begin{align} \lim_{x\to\pi}\left(\tan\left(\frac x4\right)\right)^{1/(2(x-\pi))} &=\lim_{x\to0}\left(\tan\left(\frac{x+\pi}4\right)\right)^{1/(2x)}\\[6pt] &=e^{1/4}\tag{4} \end{align} $$

Answer 5

Here, we present an approach that relies only on elementary inequalities and the squeeze theorem. To that end we begin with a primer.

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PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential functions and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)<x-1} \tag 1$$

for $x>0$

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Now, note that we can write

$$\begin{align} \lim_{x\to \pi}\left(\tan(x/4)\right)^{1/(2(x-\pi))}&=\lim_{x\to \pi}\left(1+\frac{2\tan\left(\frac{x-\pi}{4}\right)}{1-\tan\left(\frac{x-\pi}{4}\right)}\right)^{1/(2(x-\pi))}\\\\ &=\lim_{t\to 0}\left(1+\frac{2\tan(t)}{1-\tan(t)}\right)^{1/(8t)}\\\\ &=\lim_{t\to 0}e^{\frac{1}{8t}\log\left(1+\frac{2\tan(t)}{1-\tan(t)}\right)}\tag 2 \end{align}$$

Applying $(1)$ to $(2)$ reveals

$$e^{\frac{\tan(t)}{4t(1+\tan(t))}}\le e^{\frac{1}{8t}\log\left(1+\frac{2\tan(t)}{1-\tan(t)}\right)}\le e^{\frac{\tan(t)}{4t(1-\tan(t))}} \tag 3$$

whence applying the squeeze theorem to $(3)$ yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \pi}\left(\tan(x/4)\right)^{1/(2(x-\pi))}=e^{1/4}} \tag 4$$

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NOTE:

In arriving at $(4)$, we made use of the well-known limit $\lim_{x\to 0}\frac{\tan(x)}{x}=1$, which can be shown using elementary geometry only (SEE THIS ANSWER for another way forward).