Find the limit. (If the limit is infinite, enter '∞' or '-∞', as appropriate. If the limit does not otherwise exist, enter DNE.)
lim t-inf (sqrt(t)+t^2)/(9t-t^2)
the limit is -1
Apply l'Hôpital's Rule twice
Divide numerator and denominator by t^2
(sqrt(t)/t^2+t^2/t^2)/(9t/t^2-t^2/t^2) = (1/t^(3/2)+1)/(9/t -1)
lim 1/t^(3/2)=0
t->∞
lim 9/t=0
t->∞
also
lim 1/t^(3/2)=0
t-> -∞
lim 9/t=0
t-> -∞
So
lim (1/t^(3/2)+1)/(9/t -1) =
t-> -∞
(0+1)/(0-1) = 1/ - 1 = - 1
lim (1/t^(3/2)+1)/(9/t -1) =
t-> ∞
(0+1)/(0-1) = 1/ - 1 = - 1
Oh, the limit, the mathematical equivalent of going to infinity and beyond! Let's tackle this expression step by step.
We have:
lim t→∞ (sqrt(t)+t^2)/(9t-t^2)
To unravel this limit conundrum, we can divide every term by the highest power of t in the denominator, which in this case is t^2.
So, let's rewrite it as:
lim t→∞ (sqrt(t)/t^2 + t^2/t^2) / (9t/t^2 - t^2/t^2)
Now, let's simplify this expression:
lim t→∞ (1/t^(3/2) + 1) / (9/t - 1)
As t goes to infinity, the term 1/t^(3/2) will approach 0, and the term 1/t will also approach 0. Thus, we end up with:
lim t→∞ 0 + 1 / (∞ - 1)
And now, my friend, we're faced with the question of what happens when we attempt the impossible task of subtracting 1 from infinity. Since infinity is not a real number, this subtraction is undefined. So, we write:
lim t→∞ 1 / (∞ - 1) = 1 / ∞ = 0
Therefore, the limit of the expression is 0. Isn't it magical when numbers try to chase infinity and end up at zero? Remember, in the world of limits, anything is possible!
To find the limit of the given expression as t approaches negative infinity, we can simplify the expression first.
lim t→-∞ (sqrt(t) + t^2)/(9t - t^2)
We start by dividing every term in the expression by t^2, which is the highest degree term in the denominator:
lim t→-∞ (1/sqrt(t)/t^2 + t^2/t^2)/(9t/t^2 - t^2/t^2)
This simplifies to:
lim t→-∞ (1/t^2 + 1)/(9/t - 1)
Next, we can multiply the numerators and denominators by the conjugate of the denominator:
lim t→-∞ ((1/t^2 + 1)/(9/t - 1)) * ((9/t + 1)/(9/t + 1))
This gives us:
lim t→-∞ (9 + t)/(9t^3 - t + 9t^2 - t^3)
We then combine like terms in the denominator:
lim t→-∞ (9 + t)/(8t^3 + 9t^2 - t)
Since the highest degree term in the denominator is t^3, we can divide every term in the expression by t^3:
lim t→-∞ (9/t^3 + t/t^3)/(8 + 9/t - 1/t^2)
As t approaches negative infinity, 1/t becomes 0 and 1/t^2 becomes 0, leaving us with:
lim t→-∞ (0 + 0)/(8 + 0 - 0)
Finally, we get:
lim t→-∞ (0)/(8)
Therefore, the limit of the given expression as t approaches negative infinity is 0.