Before we evaluate, lets factor the numerator.

\(\displaystyle \\ \lim_{x\rightarrow 2} \frac{x^2-4x+4}{x-2} \\ \\ =\lim_{x\rightarrow 2} \frac{(x-2)(x-2)}{x-2}\)

Now lets simplify the expression to

\(\displaystyle \lim_{x\rightarrow 2} x-2\).

Now we can evaluate:

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

If we evaluate:

\(\displaystyle \lim_{x\rightarrow \pi} \frac{\cos(x)+1}{\sin(x)}\)

We get

\(\displaystyle \lim_{x\rightarrow \pi} \frac{\cos(x)+1}{\sin(x)}=\frac{0}{0}\)

So we can use L'Hopital's Rule,

L'Hopital's Rule is as follows.

If  

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{0}{0}\) 

or

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\pm\infty}{\pm\infty}\)

where a is a real number, then the following is true.

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}\). 

Now lets apply this to our problem.

\(\displaystyle \lim_{x\rightarrow \pi} \frac{\cos(x)+1}{\sin(x)}\)

\(\displaystyle =\lim_{x\rightarrow \pi} \frac{-\sin(x)}{\cos(x)}\)

\(\displaystyle =\frac{-\sin(\pi)}{\cos(\pi)}\)

\(\displaystyle =-\frac{0}{1}=0\)

If we evaluate:

\(\displaystyle \lim_{x\rightarrow 1} \frac{x^3-1}{\ln(x)}\)

We get

\(\displaystyle \lim_{x\rightarrow 1} \frac{x^3-1}{\ln(x)}=\frac{0}{0}\)

So we can use L'Hopital's Rule,

L'Hopital's Rule is as follows.

If  

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{0}{0}\) 

or

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\pm\infty}{\pm\infty}\)

where a is a real number, then the following is true.

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}\). 

Now lets apply this to our problem.

\(\displaystyle \lim_{x\rightarrow 1} \frac{x^3-1}{\ln(x)}\)

\(\displaystyle =\lim_{x\rightarrow 1} \frac{3x^2}{\frac{1}{x}}\)

\(\displaystyle =\lim_{x\rightarrow 1} 3x^3\)

Now we can evaluate the expression to get.

\(\displaystyle \lim_{x\rightarrow 1} 3x^3 =3(1)^3=3\)

If we evaluate:

\(\displaystyle \lim_{x\rightarrow 0} \frac{\ln(x+1)}{\sin(x)}\)

We get

\(\displaystyle \lim_{x\rightarrow 0} \frac{\ln(x+1)}{\sin(x)}=\frac{0}{0}\)

So we can use L'Hopital's Rule,

L'Hopital's Rule is as follows.

If  

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{0}{0}\) 

or

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\pm\infty}{\pm\infty}\)

where a is a real number, then the following is true.

\(\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}\).

Now lets apply this to our problem.

\(\displaystyle \lim_{x\rightarrow 0} \frac{\log(x+1)}{\sin(x)}\)

\(\displaystyle =\lim_{x\rightarrow 0} \frac{\frac{1}{x+1}}{\cos(x)}\)

\(\displaystyle =\lim_{x\rightarrow 0} \frac{1}{(x+1)\cos(x)}\)

If we plug in 0 we get,

\(\displaystyle \lim_{x\rightarrow 0} \frac{1}{(x+1)\cos(x)}\)

\(\displaystyle =\lim_{x\rightarrow 0} \frac{1}{(0+1)\cos(0)}=\frac{1}{1}=1\)

To evaluate the limit, we first must determine if we are approaching \(\displaystyle 3\) from the right or left. We are approaching from the left (numbers barely less than 3), and our piecewise function indicates that for numbers less than or equal to three, our function is \(\displaystyle \ln(3-x)\).

This function as it approaches zero reaches \(\displaystyle -\infty\).

Examining the graph, we want to find where the graph tends to as it approaches zero from the right hand side. We can see that there appears to be a vertical asymptote at zero. As the x values approach zero from the right the function values of the graph tend towards negative infinity.

Therefore, we can observe that \(\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=\infty\) as \(\displaystyle x\) approaches \(\displaystyle 0\)  from the right.

Examining the graph above, we need to look at three things:

1) What is the limit of the function as \(\displaystyle x\) approaches zero from the left?

2) What is the limit of the function as \(\displaystyle x\) approaches zero from the right?

3) What is the function value as \(\displaystyle x=0\) and is it the same as the result from statement one and two?

Therefore, we can observe that \(\displaystyle \lim_{x\rightarrow 0}f(x)=0\) as \(\displaystyle x\) approaches \(\displaystyle 0\)  from either side.

Examining the graph above, we need to look at three things:

1) What is the limit of the function as \(\displaystyle x\) approaches three from the left?

2) What is the limit of the function as \(\displaystyle x\) approaches three from the right?

3) What is the function value as \(\displaystyle x=3\) and is it the same as the result from statement one and two?

Therefore, we can observe that \(\displaystyle \lim_{x\rightarrow 3}f(x)\)  does not exist, as \(\displaystyle f(x)\) approaches two different limits as \(\displaystyle x\) approaches \(\displaystyle 3\): \(\displaystyle \infty\)  from the left and \(\displaystyle -\infty\) from the right.

Note that if you plug in \(\displaystyle x = 1\) to the original limit, you get \(\displaystyle \frac{0}{0}\) as your answer. This shows that you need to use L'Hopital's rule and take the derivative of both the top and the bottom of the limit and then attempt to retry finding the limit.

The derivative of the top side becomes \(\displaystyle \frac{1}{4} x^{-\frac{3}{4}}\) and the derivative of the bottom side becomes \(\displaystyle \frac{1}{2} x^{-\frac{1}{2}}\) (the 1's go away because they are constants) and so you can rewrite the problem as so: 

\(\displaystyle \lim_{x \rightarrow 1} \frac{\frac{1}{4} x^{-\frac{3}{4}}}{\frac{1}{2} x^{-\frac{1}{2}}}\).

Note that 1 raised to any power is just 1, so the limit becomes 

\(\displaystyle \frac{\frac{1}{4}}{\frac{1}{2}}\) which is \(\displaystyle \frac{1}{2}\). 

Examining the graph above, we need to look at three things:

1) What is the limit of the function as \(\displaystyle x\) approaches two from the left?

2) What is the limit of the function as \(\displaystyle x\) approaches two from the right?

3) What is the function value as \(\displaystyle x=2\) and is it the same as the result from statement one and two?

Therefore, we can observe that \(\displaystyle \lim_{x\rightarrow 2}f(x)=-6\) as \(\displaystyle x\) approaches \(\displaystyle 2\) from the left and from the right.