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Question Number 107279 by mathdave last updated on 09/Aug/20

Commented by Dwaipayan Shikari last updated on 10/Aug/20

$As x \to 0$ $tanx \to x$ $\lim_{x \to 0} (\frac{x^{x^{x^{x^{x^{x^{}}}}}}}{x^{x^{x^{x^{x^{x^{? ?}}}}}}}) = 1$
	Answered by 1549442205PVT last updated on 10/Aug/20
	$We insert the symbol for convenient I_n =lim_(n→∞) (lim_(x→0) ((((tanx)^( (tanx)^( (tanx)...) ) ^(↽^(n times) ) )/x^x^(x...) _⌣_(n times) ))) =lim_(x→0) ((f_n (x))/(g_n (x))) =where f_k (x)=(tanx)^( (tanx)^( (tan x)...) ) ^(↽^(k times) ) g_k (x)=x^x^(x....) ^(↽^(k times) ) We prove that I_n =1 for ∀n∈N,n≥1 Indeed,for n=1 we have I_1 =lim_(x→0) ((tanx)/x)=lim_(x→0) ((sinx)/x).(1/(cosx))=1.(1/1)=1 I_2 =lim_(x→0) ((tanx^(tanx) )/x^x )⇒lnI_2 =lim_(x→0) (ln((tanx^(tanx) )/x^x )) =lim_(x→0) (tanxln(tanx)−xln(x)) We have lim_(x→0) (tanxln(tanx))= J=lim_(x→0) ((lntanx)/(1/(tanx))).This is the form (∞/∞).Hence apply L′Hopital we get J=(((1+tan^2 x)/(tanx))/(−(1/(tan^2 x))(1+tan^2 x)))=lim_(x→0) (−tanx)=0 K=lim_(x→0) (xlnx)=lim_(x→0) (((lnx)/(1/x)))=lim_(x→0) ((1/x)/((−1)/x^2 )) =lim_(x→0) (−x)=0 Thus,lnI_2 =0⇒I_2 =e^0 =1.We see that the assertion is true for n=1,2.Now suppose that assertion was true for ∀n=1,2,....,k.We need to prove that it is also for n=k+1.Indeed,we conside I_(k+1) =(lim_(x→0) ((((tanx)^( (tanx)^( (tanx)...) ) ^(↽^((k+1) times) ) )/x^x^(x...) _⌣_((k+1) times) ))) Then lnI_(k+1) =lim(((tanxlnf_k (x))/(xlng_k (x)))) ⇒lnI_(k+1) =lim_(x→0) [ln(f_k (x)^(tanx) )−ln(g_k (x)^x ] +for k=1 by above proof was true +Suppose it is true ∀n=1,2,...,k which means I_k = lim_(x→0) ( ln((f_k (x))/(g_k (x))))=0 ∀k=1,2,...,k Then lnI_(k+1) =lim[ln(((f_k (x)^(tanx) )/(g_k (x)^x )))] =lim[tanxln(f_k (x))−xln(g_k (x))] =lim_(x→0) {tanx[ln(f_k (x))−ln(g_k (x))]+ tanxln(g_k (x))}=0 since [ln(f_k (x))−ln(g_k (x))]=0 by introduction the hypothesis and ln(xg_k (x))→0 when x→0.Indeed,this can is prove by induction follows as: +for k=1 we have lim_(x→0) [xlng_1 (x)]= =lim_(x→0) (xlnx)=lim_(x→0) ((lnx)/(1/x))=lim_(x→0) ((1/x)/((−1)/x^2 ))=lim_(x→0) (−x)=0 +Suppose it is true ∀n=1,2,...,k +Then lim_(x→0) [ln(g_(k+1) (x))]=lim[_(x→0) xln(g_k (x))] =0.0 due to lim[ln(g_k (x))]=0 (by introduction the hypothesis).This show that lim_(x→0) [xln(g_k (x))]=0 true ∀n≥1 Thus ,we prove that ln I_(k+1) =0 is true ,so by introduction primciple lnI_n =0 is true ∀n≥1.⇒I_n =e^0 =1∀n∈N,n≥1 (q.e.d)$
	$We insert the symbol for convenient$ $I_{n} = \lim_{n \to \infty} (\lim_{x \to 0} (\frac{\overset{\overset{n times}{↽}}{{(tanx)}^{{(tanx)}^{(tanx) . . .}}}}{{\underset{\underset{n times}{⌣}}{x}}^{x^{x . . .}}}))$ $= \lim_{x \to 0} \frac{f_{n} (x)}{g_{n} (x)}$ $= where f_{k} (x) = \overset{\overset{k times}{↽}}{{(tanx)}^{{(tanx)}^{(\tan x) . . .}}}$ $g_{k} (x) = \overset{\overset{k times}{↽}}{x^{x^{x . . . .}}}$ $We prove that I_{n} = 1 for \forall n \in N, n ⩾ 1$ $Indeed, for n = 1 we have$ $I_{1} = \lim_{x \to 0} \frac{tanx}{x} = \lim_{x \to 0} \frac{sinx}{x} . \frac{1}{cosx} = 1 . \frac{1}{1} = 1$ $I_{2} = \lim_{x \to 0} \frac{{tanx}^{tanx}}{x^{x}} \Rightarrow {lnI}_{2} = \lim_{x \to 0} (\ln \frac{{tanx}^{tanx}}{x^{x}})$ $= \lim_{x \to 0} (tanxln (tanx) - xln (x))$ $We have \lim_{x \to 0} (tanxln (tanx)) =$ $J = \lim_{x \to 0} \frac{lntanx}{\frac{1}{tanx}} . This is the form \frac{\infty}{\infty} . Hence$ $apply L^{'} Hopital we get$ $J = \frac{\frac{1 + \tan^{2} x}{tanx}}{- \frac{1}{\tan^{2} x} (1 + \tan^{2} x)} = \lim_{x \to 0} (- tanx) = 0$ $K = \lim_{x \to 0} (xlnx) = \lim_{x \to 0} (\frac{lnx}{\frac{1}{x}}) = \lim_{x \to 0} \frac{\frac{1}{x}}{\frac{- 1}{x^{2}}}$ $= \lim_{x \to 0} (- x) = 0$ $Thus, {lnI}_{2} = 0 \Rightarrow I_{2} = e^{0} = 1 . We see that$ $the assertion is true for n = 1, 2 . Now$ $suppose that assertion was true for$ $\forall n = 1, 2, . . . ., k . We need to prove that$ $it is also for n = k + 1 . Indeed, we$ $conside I_{k + 1} = (\lim_{x \to 0} (\frac{\overset{\overset{(k + 1) times}{↽}}{{(tanx)}^{{(tanx)}^{(tanx) . . .}}}}{{\underset{\underset{(k + 1) times}{⌣}}{x}}^{x^{x . . .}}}))$ $Then$ ${lnI}_{k + 1} = \lim (\frac{{tanxlnf}_{k} (x)}{{xlng}_{k} (x)})$ $\Rightarrow {lnI}_{k + 1} = \lim_{x \to 0} [\ln (f_{k} {(x)}^{tanx}) - \ln (g_{k} {(x)}^{x}]$ $+ for k = 1 by above proof was true$ $+ Suppose it is true \forall n = 1, 2, . . ., k$ $which means I_{k} = \lim_{x \to 0} (\ln \frac{f_{k} (x)}{g_{k} (x)}) = 0 \forall k = 1, 2, . . ., k$ $Then {lnI}_{k + 1} = \lim [\ln (\frac{f_{k} {(x)}^{tanx}}{g_{k} {(x)}^{x}})]$ $= \lim [tanxln (f_{k} (x)) - xln (g_{k} (x))]$ $= \lim_{x \to 0} {tanx [\ln (f_{k} (x)) - \ln (g_{k} (x))] +$ $tanxln (g_{k} (x))} = 0 since$ $[\ln (f_{k} (x)) - \ln (g_{k} (x))] = 0 by$ $introduction the hypothesis and$ $\ln ({xg}_{k} (x)) \to 0 when x \to 0 . Indeed, this can$ $is prove by induction follows as :$ $+ for k = 1 we have \lim_{x \to 0} [{xlng}_{1} (x)] =$ $= \lim_{x \to 0} (xlnx) = \lim_{x \to 0} \frac{lnx}{\frac{1}{x}} = \lim_{x \to 0} \frac{\frac{1}{x}}{\frac{- 1}{x^{2}}} = \lim_{x \to 0} (- x) = 0$ $+ Suppose it is true \forall n = 1, 2, . . ., k$ $Extra \left or missing \right$ $= 0.0 due to \lim [\ln (g_{k} (x))] = 0 (by$ $introduction the hypothesis) . This$ $show that \lim_{x \to 0} [xln (g_{k} (x))] = 0 true \forall n ⩾ 1$ $Thus, we prove that \ln I_{k + 1} = 0 is true$ $, so by introduction primciple {lnI}_{n} = 0$ $is true \forall n ⩾ 1 . \Rightarrow I_{n} = e^{0} = 1 \forall n \in N, n ⩾ 1$ $(q . e . d)$
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