anonymous
  • anonymous
help! Find the limit ,use L'Hospital's Rule to solve.limx appropriate t,(1-sinx)/cscx. How to solve?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
remember. 1/cscx = sinx and then the limit of a product is the product of the limits
anonymous
  • anonymous
yeah,I know the form about this is 0/1,and I got the directive about this ,but what's the next steps?
dumbcow
  • dumbcow
what is x approaching in the limit ?

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anonymous
  • anonymous
infinity
anonymous
  • anonymous
try convert it into \[\lim_{x \rightarrow \infty} (1-sinx) * \lim_{x \rightarrow \infty} (sinx) \] to begin with and then kick the constant (1) out the front of the limit \[(1 - \lim_{x \rightarrow \infty} (sinx)) * (\lim_{x \rightarrow \infty} (sinx) )\] and theeen - does those first steps help?
anonymous
  • anonymous
yes!
dumbcow
  • dumbcow
but sin(infinity) is indeterminate ?
anonymous
  • anonymous
right
anonymous
  • anonymous
Yes u need to either change the limit or try by rationalizing the numerator
anonymous
  • anonymous
try other method, take derivative of numerator & denominator first
anonymous
  • anonymous
I don't know how to retionalizing the numerator,get the form to 0/0,or infin/infin
anonymous
  • anonymous
to be honest i dont know if you can use l'hopitals with this one?
anonymous
  • anonymous
maybe it can't use l'hospitals
anonymous
  • anonymous
You cant use L'hopital.
anonymous
  • anonymous
It has to be 0/0 or infinity/infinity. No exceptions.
anonymous
  • anonymous
thanks,maybe it can't
anonymous
  • anonymous
sinx values oscilate between -1 and 1 and therefor you don't/can't have an indeterminate form of infinity/infinity or 0/0 in the first place with this one
anonymous
  • anonymous
^ What he said.
anonymous
  • anonymous
yeah,I think so..
anonymous
  • anonymous
Let x=(1/y) as y-->0 then x-->infinity Try using above substitutions. Although it still is a tough nut to crack :)
anonymous
  • anonymous
who?
dumbcow
  • dumbcow
my answer would be as x->infinity the function oscillates between -2 and 1/4
anonymous
  • anonymous
would you give me your steps? I want see you how to solve it
anonymous
  • anonymous
Are you sure denominator is cosec x ?
anonymous
  • anonymous
cscx,I am sure
anonymous
  • anonymous
This should help .Its a output of a limit calculator
1 Attachment
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
yes, so great!thank you very much
anonymous
  • anonymous
L'Hopital's will definitely not work here, and you'll notice this will just oscillate without any clear limit.
anonymous
  • anonymous
@Rachelzhang Anytime :)

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