To evaluate the limit as  approaches infinity, first we must factor out a term consisting of the highest power term divided by itself (equal to ):

When the factor divides to become zero, and the terms with negative exponents go to zero (when  approaches infinity), what we are left with is our final answer, .

If we plug in the value 3 directly, we get 

which is not a defined value. But we can try to factor the polynomial on the top and see if we can remove the zero at 3. As it turns out, we can:

so,

.

If we try normal substitution for this limit, we get the indeterminate form . To evaluate this limit, we use a nifty trick to bring the variable exponent down, which will give us a more accessible limit to deal with. First, suppose that the limit exists and is given as some variable , so that .

We take the natural log of both sides giving us,

 .

Since the natural log is a continuous function, we can interchange the limit symbol and the to give us,

.

Now using the formula , we can pull down the exponent and have it multiply to the , so that, 

.

Now we can try to evaluate the limit on the right hand side of the above equation. Again, we get an indeterminate form, but this time it is , which means that we can apply L'Hopital's Rule to the above limit.

Therefore we obtain:

Substituting into this limit gives us , therefore

, and so .

Solving for , we obtain .

Since 

,

then we have that

.

As t goes to infinity, the exponential in the denominator grows faster than the polynomial in the numerator, despite the fact that the polynomial is a degree higher than that in the denominator. Thus, the function goes to .

To evaluate the limit, we must examine the terms in the numerator and denominator. It is tempting to simply divide the coefficients of the  term as the answer, however in the denominator we have an exponential term that grows faster than the polynomial terms. This term reaches infinity faster and dominates the function. Thus, the denominator becomes infinitely large faster than the numerator, so the limit goes to zero.

To find the limit of this particular function at the specified x value, first factor the numerator.

After the numerator is factored, the liked terms  cancel out. Once those terms have canceled out, substitute the specified x value into the new simplified function to calculate the limit.

This question utilizes the limit definition of a derivative, which states:

Therefore, if we can define our function for our problem, we can simply evaluate this limit by taking its derivative.

  In these types of questions, you look at the second term in the numerator.  That is the negative of the function.

Since , 

We need to use the limit definition of a derivative to answer this question:

Therefore, if we can define our function for our problem, we can simply evaluate this limit by taking its derivative.

  In these types of questions, you look at the second term in the numerator.  That is the negative of the function.

Since , 

The first step is to always plug in the value of the limit. Doing so we get 

Remember, this qualifies to use l'Hôpital's Rule on the original equation. The rule says to take the derivative of both the numerator and denominator, individually. We get

This still cannot be evaluated therefore we take the derivative of both the numerator and denominator again, individually. We get

Again we cannot evaluate and must apply l'Hôpital's Rule again. We get

Which is our answer.

The first step is to always plug in the value of the limit. Doing so we get 

This qualifies to use l'Hôpital's Rule on the original equation. The rule says to take the derivative of both the numerator and denominator, individually. We get

(Note that the chain rule must be used) 

Now we can evaluate the equation. Plugging in 2 we get.

Therefore our answer is 12.