Evaluate the limit
limit t^4 − 625/t^3 − 125
t → 5
To evaluate the limit, we substitute the value of t approaching 5 into the expression t^4 - 625 / t^3 - 125:
limit t → 5 (t^4 - 625)/(t^3 - 125)
Plugging in t = 5, we get:
(5^4 - 625) / (5^3 - 125)
Simplifying further:
(625 - 625) / (125 - 125) = 0 / 0
At this point, we have an indeterminate form of 0/0, which means we need to apply further algebraic manipulation or use a different method to evaluate the limit. One method to solve this indeterminate form is by factoring and canceling common factors.
Factoring the numerator and denominator:
(t^4 - 625) = (t^2 + 25)(t^2 - 25)
(t^3 - 125) = (t - 5)(t^2 + 5t + 25)
Now we can cancel out the common factors, which are (t - 5) and (t^2 + 25):
(t^2 + 25)(t^2 - 25) / (t - 5)(t^2 + 5t + 25)
Now, plug in t = 5:
(5^2 + 25)(5^2 - 25) / (5 - 5)(5^2 + 5(5) + 25)
(25 + 25)(25 - 25) / (0)(25 + 25 + 25)
(50)(0) / (0)
Now, we have another indeterminate form of 0/0. It suggests that we apply L'Hospital's rule, which involves taking the derivative of both the numerator and the denominator with respect to t, and then repeating the process until we no longer have an indeterminate form.
Since we have already factored our expression, we can write it again as:
(t^2 + 25)(t^2 - 25) / (t - 5)(t^2 + 5t + 25)
Now, differentiate the numerator and the denominator:
Numerator:
d/dt [(t^2 + 25)(t^2 - 25)] = (2t)(t^2 - 25) + (t^2 + 25)(2t)
Denominator:
d/dt [(t - 5)(t^2 + 5t + 25)] = [(1)(t^2 + 5t + 25)] + (t - 5)(2t + 5)
Simplifying the numerator and denominator:
(2t^3 - 50t) + (2t^3 + 50t) / (t^2 + 5t + 25) + (2t^2 - 5t - 5)
Now, plug in t = 5:
(2(5)^3 - 50(5) + 2(5)^3 + 50(5)) / ((5)^2 + 5(5) + 25 + 2(5)^2 - 5(5) - 5)
Calculating further:
[2(125) - 250 + 2(125) + 250] / [25 + 25 + 25 + 2(25) - 25 - 5]
[250 - 250 + 250 + 250] / [25 + 50 - 5]
Simplifying:
500 / 70 = 7.142857...
So, the limit of the expression as t approaches 5 is approximately 7.142857...