anonymous
  • anonymous
use L'Hopital's Rule to Calculate: lim x tan (1/x) x -> infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I had got |dw:1333311842760:dw| im not sure its right tho.
roadjester
  • roadjester
You're missing the 1/x inside of the sec^2
anonymous
  • anonymous
\[\sec^2 (1/x) \] divided by \[X^2 \] is the answer?

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roadjester
  • roadjester
you have to take the limit until you get a finite answer
anonymous
  • anonymous
ok
anonymous
  • anonymous
Thanks :)
anonymous
  • anonymous
ahhh i see. ok cool!
anonymous
  • anonymous
sorry that was wrong one...
PaxPolaris
  • PaxPolaris
you can't do \[\infty \times 0.00000000001\] got to use l'hopital's rule ... for that you need to right your function as a fraction of two functions first
anonymous
  • anonymous
so its not 0?
anonymous
  • anonymous
yes
PaxPolaris
  • PaxPolaris
no not 0
PaxPolaris
  • PaxPolaris
let t = 1/x
PaxPolaris
  • PaxPolaris
\[\lim_{x \rightarrow \infty} x \cdot \tan \left( \frac 1x \right)\]\[\Large = \lim_{t \rightarrow 0}{ \tan (t) \over t}\]... now you can use l'hopital's rule and differentiate the numerator and denominator,
anonymous
  • anonymous
@PaxPolaris not quite... if you set t=1/x, then x=1/t so your limit will be \[\lim_{1/t \rightarrow \infty}\tan(t)/t\]
anonymous
  • anonymous
and x doesn't approach 0 it approaches infinity
PaxPolaris
  • PaxPolaris
t=1/x ... as x becomes big ... t becomes small
anonymous
  • anonymous
but x=1/t
PaxPolaris
  • PaxPolaris
yes so as t approches 0 .... x will become infinity
anonymous
  • anonymous
ah ok ok
PaxPolaris
  • PaxPolaris
\[\Large = \lim_{t \rightarrow 0}{\sec^2(t) \over 1} = \sec^2\left( 0 \right)=\left( 1 \over \cos(0) \right)^2=1\]
anonymous
  • anonymous
thanls PaxPolaris i get it now

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