How do you find #lim (3t^3+4)/(t^2+t-2)# as #t->1#?

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How do you find #lim (3t^3+4)/(t^2+t-2)# as #t->1#?

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Answer 1 · 2017-02-19T07:32:54

#lim_(t->1)(3t^3+4)/(t^2+t-2)=oo#

Explanation:

You simply could substitute #t=1# in #(3t^3+4)/(t^2+t-2)# and get:

#(3*1^3+4)/(1^2+1-2)=7/0=oo#

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How do you find #lim (3t^3+4)/(t^2+t-2)# as #t->1#?

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