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anonymous
 4 years ago
Is this a contradiction? On this problem they ask:
Use the definitions of the hyperbolic functions to find the following limit:
lim x>infinity sinhx
However the answer is either:
lim x>0^ coth(x) = infinity
or
lim x>0^+ coth(x) = infinity
Anyone see what I'm doing wrong? I'm attaching the original problem
anonymous
 4 years ago
Is this a contradiction? On this problem they ask: Use the definitions of the hyperbolic functions to find the following limit: lim x>infinity sinhx However the answer is either: lim x>0^ coth(x) = infinity or lim x>0^+ coth(x) = infinity Anyone see what I'm doing wrong? I'm attaching the original problem

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm not sure what you're asking...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you sure the put the right attachment

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, my bad, mistyped the uploaded file

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll take that as no one is sure what to do...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you know the standard algebraic expressions for the hyperbolic functions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x = sinh x = (e^x  e^x)/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0* sinhx = sinh x = (e^x  e^x)/2

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.2well use the definition \[\sinh (x) = (e^x  e^(x))/2\] so its the \[\lim_{x \rightarrow \infty} (e^x  e^(x))/2\] rewriting \[\lim_{x \rightarrow \infty} e^x/2  \lim_{x \rightarrow \infty} 1/(2e^x)\] 2nd part approaches 0 as x approaches infinity 1st part has approaches infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, I see, thats kind of a strange problem
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