Elementary calculus 4

Calculus Level 1

lim ⁡ t → 3 t 2 − 9 t − 3 = ? \large \lim_{t \to 3} \frac{t^2 - 9 }{t - 3} = \ ?\


The answer is 6.

Yellow Tomato by Yellow Tomato , 17, USA

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1 solution

Yellow Tomato
Nov 20, 2015

Since '3' cannot be just 'plugged in' because the denominator ends up as 0, we have to factor the fraction on the right first.

t 2 − 9 t − 3 = ( t + 3 ) ( t − 3 ) t − 3 which can be simplified to by canceling out ’t - 3 ’, to get , t + 3 \frac{t^2 - 9}{t - 3} = \frac{(t + 3)(t - 3)}{t - 3} \text{which can be simplified to by canceling out 't - 3 ', to get}, t + 3

'Plugging in t' into t + 3 , we get , t + 3 = 6 t + 3, \text {we get}, t+ 3 = 6

Answer is 6 \boxed{\boxed{6}}

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I was just about to post the exact same solution until I saw your solution.

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Aiden Lu - 5 years, 6 months ago

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You'll find that if you do more Brilliant.org, then you will find many such instances.

Krishna Karthik - 2 years, 2 months ago

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