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

Answer 1

Evelyn Agard

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]}

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Answer 2

Cameron Alexander

To find the limit of (1+2t^-1)/(7+t^-1-5t^-2) as t approaches 0, we substitute 0 for t in the expression. This gives us (1+2(0)^-1)/(7+(0)^-1-5(0)^-2). Simplifying further, we have (1+2/0)/(7+1-5/0). Since division by zero is undefined, the limit does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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