# Help with these limits!!

#### takelight

a) lim x--> 0+ (sin (x))^x

b) lim x-->infinity (pi/2 - arctan(x))^x

#### idontknow

$$\displaystyle a^b =e^{b\ln(a)}$$.

#### takelight

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

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

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#### skipjack

Forum Staff
What is $$\displaystyle \lim_{x\to \infty} (\pi /2 - \arctan(x) + 1)^{x}$$?

#### idontknow

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

Forum Staff
Think again.

2 people

#### greg1313

Forum Staff
$$\displaystyle t=\frac\pi2-\arctan x$$

$$\displaystyle x=\tan\left(\frac\pi2-t\right)$$

$$\displaystyle \lim_{t\to0}t^{\tan(\pi/2-t)}=0^1=0$$

1 person

#### idontknow

(b) $$\displaystyle 0^\infty =0$$.

#### greg1313

Forum Staff
A known basic form : $$\displaystyle 1^{\infty }$$.
That's actually an indeterminate form.