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[itex]^{}[/itex]
lim [itex]_{x\rightarrow\infty}[/itex] [itex]\left(cos\frac{1}{x}\right)[/itex][itex]^{x}[/itex]
find the limit using L' Hopital's Rule
1[itex]^{\infty}[/itex] >> change form to 0/0 by taking ln both sides
ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] x ln [itex]\left(cos\frac{1}{x}\right)[/itex]
ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] [itex]\frac{ln \left(cos\frac{1}{x}\right)}{\frac{1}{x}}[/itex] ===> 0/0
Apply L' Hopital's Rule
ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] [itex]\frac{\frac{1}{cos\frac{1}{2}}(-sin\frac{1}{x})}{\frac{-1}{x^{2}}}[/itex] ===> 0/0
the next step do i have to apply L' Hopital's Rule again or just rearrange the fraction and take exponential both sides.
please help thank you
Homework Statement
lim [itex]_{x\rightarrow\infty}[/itex] [itex]\left(cos\frac{1}{x}\right)[/itex][itex]^{x}[/itex]
find the limit using L' Hopital's Rule
Homework Equations
The Attempt at a Solution
1[itex]^{\infty}[/itex] >> change form to 0/0 by taking ln both sides
ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] x ln [itex]\left(cos\frac{1}{x}\right)[/itex]
ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] [itex]\frac{ln \left(cos\frac{1}{x}\right)}{\frac{1}{x}}[/itex] ===> 0/0
Apply L' Hopital's Rule
ln(y) = lim [itex]_{x\rightarrow\infty}[/itex] [itex]\frac{\frac{1}{cos\frac{1}{2}}(-sin\frac{1}{x})}{\frac{-1}{x^{2}}}[/itex] ===> 0/0
the next step do i have to apply L' Hopital's Rule again or just rearrange the fraction and take exponential both sides.
please help thank you