How do you find #lim (1+2t^-1)/(7+t^-1-5t^-2)# as #t->0#?

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Answer 1 · 2017-11-02T18:47:42

The answer is #0#

#lim_(t->0)(1+2/t)/(7+1/t-5/t^2)=oo/oo#

This is an indeterminate form

Let's perform some simplifications

#lim_(t->0)(1+2/t)/(7+1/t-5/t^2)=lim_(x->0)((t+2)/t)/((7t^2+t-5)/t^2)#

#=lim_(x->0)((t+2)t)/((7t^2+t-5))#

#=0/-5=0#

graph{(1+2/x)/(7+1/x-5/x^2) [-7.02, 7.024, -3.51, 3.51]}