How do you find #lim t/sqrt(4t^2+1)# as #t->-oo#?

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Answer 1 · 2016-12-12T17:25:42

#-1/2# (See below)

For #t != 0#, we have

#t/sqrt(4t^2+1) = t/sqrt(t^2(4+1/t^2))#

# = t/(sqrt(t^2)sqrt(4+1/t^2))#

We know that #sqrt(t^2) = abst#, so for #t < 0#, we have

# = t/(-t(sqrt4+1/t^2))#

# = (-1)/sqrt(4+1/t^2)#

As #trarroo#, the ration #1/t^2 rarr0#, so we get

#lim_(xrarr-oo) t/sqrt(4t^2+1) = lim_(xrarr-oo) (-1)/sqrt(4+1/t^2)#

# = (-1)/sqrt(4+0) = -1/2#