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

1 Answer
Steve M
Dec 2, 2016

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

Explanation:

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 #

We can confirm this with a graph:
graph{(2x+1)^4/(3x^2+1) [-3.208, 2.952, -0.11, 2.968]}

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