T takelight Nov 2019 7 0 California Nov 13, 2019 #1 a) lim x--> 0+ (sin (x))^x b) lim x-->infinity (pi/2 - arctan(x))^x

T takelight Nov 2019 7 0 California Nov 13, 2019 #3 ok I see how the expression you gave can solve the first problem. However, if I do it for the second one, I'm still stuck with ln(pi/2-arctan(x)) and this would give me ln(0) which is -infinity???

ok I see how the expression you gave can solve the first problem. However, if I do it for the second one, I'm still stuck with ln(pi/2-arctan(x)) and this would give me ln(0) which is -infinity???

idontknow Dec 2015 972 128 Earth Nov 13, 2019 #4 \(\displaystyle \lim_{x\rightarrow \infty} (\pi /2 - \arctan(x))^{x} =\lim_{x\rightarrow \infty} (1+[\pi /2 - \arctan(x)-1])^{\displaystyle x\cdot \displaystyle \frac{\pi/2 - \arctan(x)-1}{\pi/2 - \arctan(x)-1}}=\lim_{x\rightarrow \infty} e^{x(\pi /2 -\arctan(x)-1)}=\lim_{x\rightarrow \infty}e^{-x}=0\). Last edited by a moderator: Nov 14, 2019

\(\displaystyle \lim_{x\rightarrow \infty} (\pi /2 - \arctan(x))^{x} =\lim_{x\rightarrow \infty} (1+[\pi /2 - \arctan(x)-1])^{\displaystyle x\cdot \displaystyle \frac{\pi/2 - \arctan(x)-1}{\pi/2 - \arctan(x)-1}}=\lim_{x\rightarrow \infty} e^{x(\pi /2 -\arctan(x)-1)}=\lim_{x\rightarrow \infty}e^{-x}=0\).

skipjack Forum Staff Dec 2006 21,379 2,410 Nov 14, 2019 #5 What is \(\displaystyle \lim_{x\to \infty} (\pi /2 - \arctan(x) + 1)^{x}\)?

idontknow Dec 2015 972 128 Earth Nov 14, 2019 #6 skipjack said: What is \(\displaystyle \lim_{x\to \infty} (\pi /2 - \arctan(x) + 1)^{x}\)? Click to expand... A known basic form : \(\displaystyle 1^{\infty }\).

skipjack said: What is \(\displaystyle \lim_{x\to \infty} (\pi /2 - \arctan(x) + 1)^{x}\)? Click to expand... A known basic form : \(\displaystyle 1^{\infty }\).

greg1313 Forum Staff Oct 2008 8,008 1,174 London, Ontario, Canada - The Forest City Nov 14, 2019 #8 \(\displaystyle t=\frac\pi2-\arctan x\) \(\displaystyle x=\tan\left(\frac\pi2-t\right)\) \(\displaystyle \lim_{t\to0}t^{\tan(\pi/2-t)}=0^1=0\) Reactions: 1 person

\(\displaystyle t=\frac\pi2-\arctan x\) \(\displaystyle x=\tan\left(\frac\pi2-t\right)\) \(\displaystyle \lim_{t\to0}t^{\tan(\pi/2-t)}=0^1=0\)

greg1313 Forum Staff Oct 2008 8,008 1,174 London, Ontario, Canada - The Forest City Nov 17, 2019 #10 idontknow said: A known basic form : \(\displaystyle 1^{\infty }\). Click to expand... That's actually an indeterminate form.

idontknow said: A known basic form : \(\displaystyle 1^{\infty }\). Click to expand... That's actually an indeterminate form.