Let's start by finding the slope of a secant line that passes through the points \((0,0)\) and \((1,1)\).

\[m_{sec}= {{1-0} \over {1-0}}=1\]

Now to find the tangent line, we want the other point (other than \((1,1)\)) to get infinitely close to \((1,1)\) so that the secant line will resemble the tangent line. therefore; let's replace the pont \((0,0)\) with a point the represents any point on the graph: \((a,a^2)\). Let's now find the secant line with this universal point.

\[m_{sec}= {{1-a^2} \over {1-a}}\]

Let's simplify the above expression.

\[m_{sec}= {{(1-a)(1+a)} \over {1-a}}=(1+a)\]

Let's set up a table to determine the value of the limit as \(a\) approaches \(1\)

\begin{array} {|r|r|}\hline \textbf a &  \textbf m_{sec} \\ \hline 0 & 1 \\ \hline 0.75 & 1.75 \\ \hline 0.85 & 1.85 \\ \hline 0.99 & 1.99 \\ \hline 1 & DNE \\ \hline 1.01 & 1.01 \\ \hline 1.15 & 1.15 \\ \hline 1.25 & 1.25 \\ \hline 2 & 3\\ \hline  \end{array}

From looking at the table, \(m_{sec}\) approaches \(2\) as \(a\) approaches \(1\) from either side (\(a>1 \: and \: a

\[m_{tan}=2\]

Let's start by finding the slope of a secant line that passes through the points \((0,0)\) and \((1,1)\).

\[m_{sec}= {{1-0} \over {1-0}}=1\]

Now to find the tangent line, we want the other point (other than \((1,1)\)) to get infinitely close to \((1,1)\) so that the secant line will resemble the tangent line. therefore; let's replace the pont \((0,0)\) with a point the represents any point on the graph: \((a,a^3)\). Let's now find the secant line with this universal point.

\[m_{sec}= {{1-a^3} \over {1-a}}\]

Let's simplify the above expression.

\[m_{sec}= {{(a-1)(a^2+a+1)} \over {a-1}}\]

Let's set up a table to determine the value of the limit as \(a\) approaches \(1\)

\begin{array} {|r|r|}\hline \textbf a &  \textbf m_{sec} \\ \hline 0 & 1 \\ \hline 0.75 & 2.3125 \\ \hline 0.85 & 2.5725 \\ \hline 0.99 & 2.9701 \\ \hline 1 & DNE \\ \hline 1.01 & 3.0301 \\ \hline 1.15 & 3.4275 \\ \hline 1.25 & 3.8125 \\ \hline 2 & 7 \\ \hline  \end{array}

From looking at the table, \(m_{sec}\) approaches \(3\) as \(a\) approaches \(1\) from either side (\(a>1 \: and \: a

\[m_{tan}=3\]

As we can see from the graph above, if we approach \(0\) from the right, the function approaches \(1\). If we approach \(0\) from the right, the function also approaches \(1\), and since both the right and left sided limits equal each other, the limit exists and is equal to \(1\)

This can be written as

\[\mathop {\lim }\limits_{x \to 0} \cos{(x)} =1.\]

Let’s start with 1. If we approach one from the right, we can see the graph approach an asymptote and tend towards positive infinity; therefore,

\[\mathop {\lim }\limits_{x \to 1^+} f\left( x \right)=+\infty. \]

Next, if we approach one from the left, we can see the graph approach the same asymptote, except the graph tends towards negative infinity; therefore,

\[\mathop {\lim }\limits_{x \to 1^-} f\left( x \right)=-\infty. \]

However, since both the left sided limit and the right sided limit do not equal each other, the limit as x approaches one does not exist.

\[\mathop {\lim }\limits_{x \to 1} f\left( x \right)=DNE. \]

Next, if we approach two from the right, we can see the graph tend toward a removable discontinuity at x=2; therefore,

\[\mathop {\lim }\limits_{x \to 2^+} f\left( x \right)=2. \]

Next, if we approach two from the left, we can see the graph tend toward the same removable discontinuity at x=2; therefore,

\[\mathop {\lim }\limits_{x \to 2^-} f\left( x \right)=2. \]

Also, since both the left and right sided limits equal each other, we can say that the limit exists and that the limit is equal to two.

\[\mathop {\lim }\limits_{x \to 2} f\left( x \right)=2 \]

Next, if we approach three from either side, we can see the graph approach negative one; therefore, we can say that  

\[\mathop {\lim }\limits_{x \to 3^+} f\left( x \right)=-1, \]

\[\mathop {\lim }\limits_{x \to 3^-} f\left( x \right)=-1, \]

and finally that

\[\mathop {\lim }\limits_{x \to 3} f\left( x \right)=-1. \]

Lastly, if we approach five from the right, the graph approaches three, but if we approach five from the left, the graph approaches 4; therefore, we can conclude that

\[\mathop {\lim }\limits_{x \to 5^+} f\left( x \right)=3, \]

\[\mathop {\lim }\limits_{x \to 5^-} f\left( x \right)=4, \]

and finally that

\[\mathop {\lim }\limits_{x \to 5} f\left( x \right)=DNE. \]

The first thing we should try when we see a limit is to substitute the value we are approaching into the function.

1. If we substitute \(-4\) into the function, we will get that the function approaches \(1\) \[\therefore\] \[\mathop {\lim }\limits_{x \to -4} (x^{2}+10x+25)=1.\]
2. If we substitute \(\frac{\pi}{2}\) into the function, we will get that the function approaches \(0\) \[\therefore\] \[\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\cos{(x)}}{x}=0.\]
3. If we substitute \(0\) into the function, we will get that the function approaches the undefined value of \(\frac{1}{0}\); therefore, we need to evaluate the right and left sided limits. Let's evaluate the left-sided limit first. If we subsite several values that are both left of \(0\) and close to the value \(0\), we can see that the function approaches \(-\infty\). \begin{array} {|r|r|}\hline x & \frac{1}{2x^{2}+2x} \\ \hline -0.5 & -2 \\ \hline -0.25 & -2.667 \\ \hline -0.125 & -4.571 \\ \hline -0.0625 & -8.533 \\ \hline -0.03125 & -16.516 \\ \hline 0 & undef \\ \hline  \end{array} \[\therefore\] \[\mathop {\lim }\limits_{x \to 0^-}\frac{1}{2x^{2}+2x}=-\infty\]Next, let's evaluate the right-sided limit by substituting in several values that are both close to \(0\) and the right of \(0\). If we do that, we can see that the function approaches \(\infty\). \begin{array} {|r|r|}\hline x & \frac{1}{2x^{2}+2x} \\ \hline 0.5 & 0.667 \\ \hline 0.25 & 1.6 \\ \hline 0.125 & 3.556 \\ \hline 0.0625 & 7.529 \\ \hline 0.03125 & 15.515 \\ \hline 0 & undef \\ \hline  \end{array} \[\therefore\] \[\mathop {\lim }\limits_{x \to 0^+}\frac{1}{2x^{2}+2x}=\infty\]Since the two one-sided limits do not equal each other, the limit itself does not exist. \[\mathop {\lim }\limits_{x \to 0}\frac{1}{2x^{2}+2x}=DNE\]
4. If we substitute \(0\) into the function, we will get that the function approaches the undefined value of \(\frac{4}{0}\); therefore, we need to evaluate the right and left sided limits. Let's evaluate the left-sided limit first. If we subsite several values that are both left of \(0\) and close to the value \(0\), we can see that the function approaches \(+\infty\). \begin{array} {|r|r|}\hline x & \frac{4}{x^{2}} \\ \hline -1 & 4 \\ \hline -0.5 & 16 \\ \hline -0.25 & 64 \\ \hline -0.125 & 256 \\ \hline -0.0625 & 1024 \\ \hline 0 & undef \\ \hline  \end{array} \[\therefore\] \[\mathop {\lim }\limits_{x \to 0^-}\frac{4}{x^{2}}=\infty\]Next, let's evaluate the right-sided limit by substituting in several values that are both close to \(0\) and the right of \(0\). If we do that, we can see that the function approaches \(\infty\). \begin{array} {|r|r|}\hline x & \frac{4}{x^{2}} \\ \hline 1 & 4 \\ \hline 0.5 & 16 \\ \hline 0.25 & 64 \\ \hline 0.125 & 256 \\ \hline 0.0625 & 1024 \\ \hline 0 & undef \\ \hline  \end{array} \[\therefore\] \[\mathop {\lim }\limits_{x \to 0^+}\frac{4}{x^{2}}=\infty\] Since the two one sided limits both approach \(\infty\) the limit exists and approaches \(\infty\). \[\mathop {\lim }\limits_{x \to 0}\frac{4}{x^{2}}=\infty\]

1. From looking at this limit, we can see that we can not substitute \(0\) into the function because it will lead to an indeterminate form; therefore, let's jump straight to analyzing the right and left hand limits. To evaluate the right-sided limit, let's substitute \(0.00000001\), a value very close to \(0\) and right of \(0\), using our calculator. \[\frac{\sin{(0.0000001)}}{0.00000001}\] If we do that, we get \(1\). If we do the same for the left, except with \(-0.00000001\) instead, we will also get \(1\). \[\therefore\] \[\mathop {\lim }\limits_{x \to 0}\frac{\sin{(x)}}{x}=1\] Memorizing this limit is beneficial as it will help solve other limits. One example is the following limit.
2. Once again, we can see that we can not substitute \(0\) into the function because it will lead to an indeterminate form; however, instead of solving the left and right hand limits we can use algebra in our favor to simplify this problem. Let's algebraically rewrite this problem. \[\mathop {\lim }\limits_{x \to 0}\frac{\tan{(x)}}{x}=\mathop {\lim }\limits_{x \to 0}\bigg(\frac{\sin{(x)}}{\cos{(x)}}\bigg)\bigg(\frac{1}{x}\bigg)\] \[=\mathop {\lim }\limits_{x \to 0}\bigg(\frac{\sin{(x)}}{x}\bigg)\bigg(\frac{1}{\cos{(x)}}\bigg)\] We know the value of \(\mathop {\lim }\limits_{x \to 0}\frac{\sin{(x)}}{x}\) is \(1\); therefore, we can substitute that in. and we only have to evaluate \[\mathop {\lim }\limits_{x \to 0}\bigg(\frac{1}{\cos{(x)}}\bigg).\] To solve the above limit we can look at the right and left hand limits. If we substitute in \(0.00000001\) and \(-0.00000001\), the value of the limit will be \(1\). \[\therefore\] \[\mathop {\lim }\limits_{x \to 0}\frac{\tan{(x)}}{x}=1\]