How do you find the limit of #((t^2)+(5t)) / (cosh(t)-1)# as t approaches 0?

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Answer 1 · 2016-07-12T06:07:01

#lim_{t to 0^-} (t^2+5t) / (cosh(t)-1) = -oo#

#lim_{t to 0^+} (t^2+5t) / (cosh(t)-1) = oo#

#lim_{t to 0} (t^2+5t) / (cosh(t)-1)#

currently this is #0/0#, indeterminate, so we can apply Lhopital's Rule

#= lim_{t to 0} (2t+5) / (sinh(t))#

#= lim_{t to 0} 2 (t) / (sinh(t)) +5 / (sinh(t)#

From L'Hopital: # lim_{t to 0} 2 (t) / (sinh(t)) = 2 lim_{t to 0} 1 / (cosh(t)) = 2#

#implies 2 + lim_{t to 0} 5 / (sinh(t)#

Now, # lim_{t to 0} 5 / (sinh(t)# is 2 sided limit ie

#= lim_{t to 0^-} 5 / sinh(t) = - oo#

#= lim_{t to 0^+} 5 / sinh(t) = oo#

so

#lim_{t to 0^-} (t^2+5t) / (cosh(t)-1) = -oo#

#lim_{t to 0^+} (t^2+5t) / (cosh(t)-1) = oo#