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How do you find #lim (5+6t^-1+t^-2)/(8-3t^-1+2t^-2)# as #t->0#?

Answer 1

Samantha Adkison

#1/2#

As the power is negative that it is hard to take the limit, we multiply by #t^2/t^2# to cancel out the negative power.

#lim_(t->0) (5+6t^(-1)+t^(-2))/(8-t^(-1)+2t^(-2))* t^2/t^2#

#=lim_(t->0) (5t^2+6t+1)/(8t^2-t+2#

#=(5(0)^2+6(0)+1)/(8(0)^2-(0)+2#

#=1/2#

Here is the answer :)

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Answer 2

Emma Aldridge

To find the limit of the given expression as t approaches 0, we substitute 0 for t in the expression and simplify:

lim (5+6t^-1+t^-2)/(8-3t^-1+2t^-2) as t->0

Substituting t=0:

(5+6(0)^-1+(0)^-2)/(8-3(0)^-1+2(0)^-2)

Simplifying:

(5+6(0)+(0))/(8-3(0)+2(0))

(5+0+0)/(8-0+0)

5/8

Therefore, the limit of the given expression as t approaches 0 is 5/8.

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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 (5+6t^-1+t^-2)/(8-3t^-1+2t^-2)# as #t-&gt;0#?

How do you find #lim (5+6t^-1+t^-2)/(8-3t^-1+2t^-2)# as #t->0#?