In general, when you are looking for \(\displaystyle \lim_{x \rightarrow a^+} y\) you are looking to see whether the limit of y exists to the right, and if it does, what is the value.

Solution:

In this case, we want to see the limit at \(\displaystyle x=11\), from the right. The limit exists, and it corresponds to the function \(\displaystyle -6\). Since this function has no inputs, the limit is \(\displaystyle -6\).

As you go to \(\displaystyle x=0\) from the left, what value do you get closer to?

The limit of \(\displaystyle x=0\) from the left is 

\(\displaystyle \frac{0^2 + 2}{0-2} = -1\).

To evaluate the limit, we must first see whether the limit is right or left sided. The negative sign "exponent" on \(\displaystyle \pi\) indicates that we are evaluating the limit from the left side, using \(\displaystyle x\) values slightly less than \(\displaystyle \pi\). So, the part of the piecewise function we will use is the first one; when we evaluate the limit, we get an answer of \(\displaystyle -1\) 

(\(\displaystyle \cos(\pi)=-1\)).

In general, when you are looking for \(\displaystyle \lim_{x \rightarrow a^+} y\) you are looking to see whether the limit of y exists to the right, and if it does, what is the value.

Solution:

In this case, we want to see the limit at \(\displaystyle x=9\) , from the right. The limit exists, and the value corresponds to the function \(\displaystyle x^2+9\)

\(\displaystyle x^2+9 = 9^2 + 9 = 81 + 9 = 90\)

In general, when you are looking for \(\displaystyle \lim_{x \rightarrow a^-} y\) you are looking to see whether the limit of y exists to the left, and if it does, what is the value.

Solution:

In this case, we want to see the limit at \(\displaystyle x=7\), from the left. The limit exists, and the value corresponds to the function \(\displaystyle x-5\)

\(\displaystyle x-5 = 7-5 = 2\)

In general, when you are looking for \(\displaystyle \lim_{x \rightarrow a^-} y\) you are looking to see whether the limit of y exists to the left, and if it does, what is the value.

Solution:

In this case, we want to see the limit at \(\displaystyle x=3\), from the left. The limit exists, and the value correponds to the function \(\displaystyle x-4\)

\(\displaystyle x-4 = 3-4 = -1\)

To evaluate the limit, we must first determine whether the limit is right or left sided. The positive sign "exponent" on 7 indicates that we are evaluating the limit from the right side, or using numbers slightly larger than 7. The part of the piecewise function corresponding to these values is the second function; when we evaluate the limit using that function, we approach \(\displaystyle -\infty\) (the natural log function approaches negative infinity as x approaches zero).

The limit exists if 

 \(\displaystyle \lim_{x\rightarrow a^-}f(x) = \lim_{x\rightarrow a^+}f(x)\)

because 

 \(\displaystyle f(1.9)=1.9-2=-0.1\)

 \(\displaystyle f(1.99)=1.99-2=-0.01\)

 \(\displaystyle f(1.999)=1.999-2=-0.001\)

we see that 

 \(\displaystyle \lim_{x\rightarrow 2^-}f(x) = 0\)

because 

\(\displaystyle f(2.1)=2.1-2=0.1\) 

 \(\displaystyle f(2.01)=2.01-2=0.01\)

 \(\displaystyle f(2.001)=2.001-2=0.001\)

we see that 

\(\displaystyle \lim_{x\rightarrow 2^+}f(x) = 0\) 

since 

\(\displaystyle \lim_{x\rightarrow a^-}f(x) = \lim_{x\rightarrow a^+}f(x)=0\) 

we conclude that the limit exists and 

\(\displaystyle \lim_{x\rightarrow 2}f(x) = 0\)

To evaluate the limit, we must make sure that the same value is being approached from both sides. When we evaluate the limit from the left (the first part of the piecewise function) and the right (the second part of the piecewise function), we get the same value (\(\displaystyle 7\)) so the limit is equal to \(\displaystyle 7\).

To evaluate the limit, we must first determine whether the limit is right or left sided; the positive sign "exponent" on the \(\displaystyle 2\) indicates that the limit is right sided, or that we are approaching \(\displaystyle 2\) with values slightly greater than \(\displaystyle 2\). This corresponds to the second half of the piecewise function, and when we evaluate the limit using that function we get \(\displaystyle \ln(2)\).