There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

For the limit of  we cannot use l'Hopital's rule or any of the other standard tricks, because sin(x) itself does not have a limit at infinity. We have to use the Squeeze Theorem.

The first thing we can notice is that no matter what x is, 

. If we divide all parts of the inequality by , we get

so our limit must be between  and .

We know both of these limits are zero, because  increases without bound, so its inverse must go to 0. Therefore we get

so 

To determine the limit of the piecewise function, we must determine whether the limit is the same approaching zero from the right or left of zero. From the right (values smaller than zero), we approach 1, and from the left of zero (values greater than zero), we approach 1 as well. Thus, the limit is 1.

To evaluate the limit as x approaches infinity, we must factor out a term consisting of the highest power term divided by itself (so we are unchanging the limit):

The first term cancels to become 1, and the negative exponent terms go to zero when x approaches infinity (infinity in the denominator makes zero), so what we are left with is 

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=1; so we proceed to insert the value of x into the entire equation.

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=2; so we proceed to insert the value of x into the entire equation.

We see that we cannot factor this to make the denominator not equal 0; hence this limit DNE because the denominator is zero.

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=5; so we proceed to insert the value of x into the entire equation.

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=6; so we proceed to insert the value of x into the entire equation.

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=7; so we proceed to insert the value of x into the entire equation.