What is the limit as t approaches 2 [2-1/t^2-4t+4]?
回答 (3)
2018-02-26 5:49 am
That depends on what the expression in brackets is. It is impossible to tell without parentheses.
Is it 2 - (1/t^2) - 4t + 4?
Is it (2 - 1)/(t^2 - 4t + 4)?
Is it 2 - [1/(t^2 - 4t + 4)] ?
Is it 2 - [1/(t^2 - 4t)] + 4?
Is it something else?
What's in the fraction? What's in the numerator? What's in the denominator?
Is it 2 - (1/t^2) - 4t + 4?
Is it (2 - 1)/(t^2 - 4t + 4)?
Is it 2 - [1/(t^2 - 4t + 4)] ?
Is it 2 - [1/(t^2 - 4t)] + 4?
Is it something else?
What's in the fraction? What's in the numerator? What's in the denominator?
2018-02-26 7:52 am
Take derivative
2-(1/(t²-4t+4)
2t-4/(t-4t+4)²
2/(t-2)³
Plug in 2
2/0------> limit is -inf.
2-(1/(t²-4t+4)
2t-4/(t-4t+4)²
2/(t-2)³
Plug in 2
2/0------> limit is -inf.
2018-02-26 6:19 am
Is this:
2 - 1 / (t^2 - 4t + 4)?
1 / (t^2 - 4t + 4) =>
1 / (t - 2)^2
As t approaches 2 , 1 / (t - 2)^2 approaches infinity
2 - inf = -inf
The limit is negative infinity
2 - 1 / (t^2 - 4t + 4)?
1 / (t^2 - 4t + 4) =>
1 / (t - 2)^2
As t approaches 2 , 1 / (t - 2)^2 approaches infinity
2 - inf = -inf
The limit is negative infinity
收錄日期: 2021-04-24 00:59:26
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