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How do you find the limit #lim_(t->-3)(t^2-9)/(2t^2+7t+3)# ?

Answer 1

Scarlett Allison

#lim_{t to -3}{t^2-9}/{2t^2+7t+3}#

by factoring out the numerator and the denominator,

#=lim_{t to -3}{(t+3)(t-3)}/{(t+3)(2t+1)}#

by cancelling out #(t-3)#'s,

#=lim_{t to -3}{t-3}/{2t+1}={(-3)-3}/{2(-3)+1}={-6}/{-5}=6/5#

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Answer 2

Paisley Johnson

To find the limit of the given expression, we can substitute the value of t approaching -3 into the expression and simplify it. By substituting -3 into the expression, we get (-3^2 - 9) / (2(-3)^2 + 7(-3) + 3). Simplifying further, we have (9 - 9) / (18 - 21 + 3). Continuing to simplify, we get 0 / 0. This is an indeterminate form. To evaluate this limit, we can use algebraic manipulation or L'Hôpital's rule.

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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 the limit #lim_(t-&gt;-3)(t^2-9)/(2t^2+7t+3)# ?

How do you find the limit #lim_(t->-3)(t^2-9)/(2t^2+7t+3)# ?