Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to1}{\frac{|x-1|}{x-1}}}\)

\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)

Okay, so first if we try to just evaluate the limit at \(x=5\), we get \(6/0\). Our first inclination might be to just say the limit is positive infinity. However \(6/0\) doesn't give us enough information to say whether the limit goes to positive infinity or negative infinity. We need to know the sign of the denominator.
As the denominator gets smaller and smaller, the magnitude of the fraction gets larger and larger but, since the numerator is positive, the sign of denominator will determine whether the limit goes to positive or negative infinity.

Here is the logic we use to determine the limit. Since \(x\) is approaching \(5\) from the right, \(x\) will always be larger than \(5\), so \(x-5\) is always positive. To check this, let's test a few numbers.
\(x=5.1 \to 6/(5.1-5) = 6/0.1 = 60 \gt 0 \)
\(x = 5.01 \to 6/(5.01-5) = 6/0.01 = 600 \gt 0\)

So the test values go along with our argument about the denominator always being positive. And since the numerator is also always positive, then the resulting fraction will always be positive. So now we can confidently say that the limit goes to positive infinity.

Although we were not asked to in the problem, we decided to plot the function to check our answer. Here is the plot.

\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\) \( = + \infty \)

\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)

Okay, so first if we try to just evaluate the limit at \(x=0\), we get \(-1/0\). Our first inclination might be to just say the limit is negative infinity. However \(-1/0\) doesn't give us enough information to say whether the limit goes to positive infinity or negative infinity. We need to know the signs of both the numerator and denominator on both sides of zero.
First, let's look at the denominator. Of course, the \(x^2\) term is always positive. Since \(x\) is very close to zero, \(x+2\) also always positive. This is the case whether \(x\) is positive or negative. Consequently, the sign of the numerator will determine the limit.

Here is the logic we use to determine the limit. We need to look at each side of \(x=0\) separately.

For \(x \lt 0 \), \(x-1\) is negative. Similarly for \(x \gt 0\), for \(x\) very close to zero, \(x-1\) is also negative. In equation form, \(\displaystyle{ \lim_{x\to0^-}{ \frac{x-1}{x^2(x+2)} } = - \infty }\) and \(\displaystyle{ \lim_{x\to0^+}{ \frac{x-1}{x^2(x+2)} } = - \infty }\). Since the limit from the left is equal to the limit from the right, we know that \(\displaystyle{ \lim_{x\to0}{ \frac{x-1}{x^2(x+2)} } = - \infty }\).

Although we were not asked to in the problem, we decided to plot the function to check our answer. Here is the plot.

\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\) \( = - \infty \)

Evaluate the following limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to-4^-}{|x+4|}}\)

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^+}{\sqrt{x^2-25}}}\)

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^-}{\sqrt{x(5-x)}}}\)

Evaluate the limit \(\displaystyle{ \lim_{x\to0}{1/x} }\) giving your answer in exact form.

For this solution, we will look at specific values of \(x\) to give us a feel for what is happening. Although not recommended for most problems, this technique, along with the graph, can give a good understanding of this function around \(x=0\). Of course, we can't just plug in zero for \(x\) since we end up with \(1/0\) which is not valid.

Here is the plot of \(1/x\).

Since the limits are different on each side of the graph, i.e. \(\displaystyle{ \lim_{x\to0^+}{1/x} \neq \lim_{x\to0^-}{1/x} }\), the limit \(\displaystyle{\lim_{x\to0}{1/x}}\) does not exist.

The limit does not exist.

Evaluate the limit \(\displaystyle{ \lim_{x\to0}{1/x^2} }\) giving your answer in exact form.

For this solution, we will look at specific values of \(x\) to give us a feel for what is happening. Although not recommended for most problems, this technique, along with the graph, can give a good understanding of this function around \(x=0\). Of course, we can't just plug in zero for \(x\) since we end up with \(1/0\) which is not valid.

Here is the plot of \(1/x^2\).

Since \(\displaystyle{ \lim_{x\to0^+}{1/x} = \lim_{x\to0^-}{1/x} }\), the limit \(\displaystyle{\lim_{x\to0}{1/x} = \infty}\).

\(\displaystyle{ \lim_{x\to0}{1/x^2} }\) \( = \infty \)

Evaluate \(\displaystyle{\lim_{x\to1}{f(x)}}\) for \(\displaystyle{f(x)=\left\{\begin{array}{lr} x+3 & x \leq 1 \\ x^2-2x & x >1 \end{array}\right.}\)

Prove that the limit \(\displaystyle{\lim_{x\to0}{\frac{|x|}{x}}}\) does not exist.

Evaluate the limit \(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\) giving your answer in exact form.

This is a quite interesting function. First, let's look at the equation. Since the function \(\displaystyle{ f(x) = \frac{x-2\abs{x}}{\abs{x}} }\) contains absolute values, we need to set up a piecewise function.

When \(x \lt 0\), \(\displaystyle{ f(x) = \frac{x+2x}{-x} = -3 }\)

When \(x \gt 0\), \(\displaystyle{ f(x) = \frac{x-2x}{x} = -1 }\)

So our piecewise function is
\( f(x) = \left\{ \begin{array}{rrl} -3 & & x \lt 0 \\ -1 & & x \gt 0 \end{array} \right. \)

Here is the plot of \(f(x)\).

So we can see from the piecewise function and the plot that \(\displaystyle{\lim_{x\to0^+}{f(x)} = -1 }\) and \(\displaystyle{\lim_{x\to0^-}{f(x)} = -3 }\). Since the limits are different on each side of \(x=0\), the limit does not exist.

The limit does not exist.