Question

Compute limit Help?

Answer 1

for part i)\[\lim_{x \rightarrow \infty} 5^{\frac{ -1 }{ x }}\cos(\frac{ \pi }{ 6x })= 5^{-0}\cos0=1\]

part ii) \[-1 \le \cos (\frac{ \pi }{ 6x }) \le1\]
\[-5^{\frac{ -1 }{ x }} \le 5^{\frac{ -1 }{ x }}\cos(\frac{ \pi }{ 6x }) \le 5^{\frac{ -1 }{ x }}\]
lim lwft is not equal to lim right? then how?

Answer 2

part 2 only asks for the right hand limit

Answer 3

hmm, but if i plug in 0 , there will be cos infinity? which i cant evaluate

Answer 4

you can still use the squeeze theorem as you were doing earlier. the left and right hand limits are actually the same

Answer 5

isit.. so mean i just ignore the sign of 5?

Answer 6

what's\[\lim_{x\to 0}a^{-1/x}\]?

Answer 7

a^-ve infty?

Answer 8

right, which is what?

Answer 9

the limit at the right

Answer 10

\[a^{-b}=\frac1{a^b}\]

Answer 11

\[a^{-\infty}=\frac1{a^{\infty}}=?\]

Answer 12

0?

Answer 13

yes

Answer 14

okay that make sense now thank you! :D

Answer 15

welcome!