How do you find the limit of #(2t+1)^4/(3t^2+1)^2# as #t->0#?

Answer 1

# lim_(t rarr 0) (2t+1)^4/(3t^2+1)^2 = 1 #

Both the numerator and denominator are continuous over all real numbers. The function does not have any singularities and the denominator is positive for all real numbers, hence we can just substitute #t=0# to get:
# lim_(t rarr 0) (2t+1)^4/(3t^2+1)^2 = (0+1)^4/(0+1)^2 = 1 #

A graph such as graph{(2x+1)^4/(3x^2+1) [-3.208, 2.952, -0.11, 2.968]} allows us to verify this.

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Answer 2

To find the limit of (2t+1)^4/(3t^2+1)^2 as t approaches 0, we can substitute 0 for t in the expression. This gives us (2(0)+1)^4/(3(0)^2+1)^2, which simplifies to 1^4/1^2. Therefore, the limit is equal to 1.

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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.

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.

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.

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.

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