Indeterminate Forms

Consider the following limit:

\[ \lim_{x \to 4} \frac{x^2-16}{x-4}\]

If you try to substitute \(x\) for \(4\) in the above limit, you will find that:

\[ \begin{align} \lim_{x \to 4} \frac{x^2-16}{x-4} &= \frac{4^2-16}{4-4} \\ &= \frac{16-16}{4-4} \\ &= \frac{0}{0} \end{align}\]

which is an indeterminate form of:

\[ \frac{0}{0}\]

To properly evaluate this limit, you can factor the difference of squares, so you can cancel the like terms, that is:

\[ \begin{align} \lim_{x \to 4} \frac{x^2-16}{x-4} &= \lim_{x \to 4} \frac{(x+4)\cancel{(x-4)}}{\cancel{(x-4)}} \\ &= \lim_{x \to 4} (x+4) \\ &= 4+4 \\&= 8\end{align}\]

In the previous example, you evaluated the limit:

\[ \lim_{x \to 4} \frac{x^2-16}{x-4}\]

By factorizing the numerator. If this particular factorization does not come to your mind, you can also use L'Hôpital's rule, obtaining:

\[ \begin{align} \lim_{x \to 4} \frac{x^2-16}{x-4} &= \lim_{x \to 4} \frac{2x}{1} \\ &= \frac{2(4)}{1} \\ &= 8\end{align} \]

Evaluate the limit:

\[ \lim_{x \to 0^+} \left( \frac{1}{x}-\frac{1}{x^2} \right)\]

Solution:

If you were to evaluate the limit by direct substitution, you would find that:

\[ \lim_{ x \to 0^+} \left( \frac{1}{x} - \frac{1}{x^2}\right)= \infty - \infty\]

So this is an indeterminate form.

Instead of evaluating directly, try subtracting both fractions, that is:

\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{x^2} \right)= \lim_{x \to 0^+} \left( \frac{x-1}{x^2}\right)\]

So you can inspect the limit by direct substitution.

The numerator is negative, and the denominator is positive as you approach \(0\) from the right, so the result will be negative infinity, that is:

• An indeterminate form is an expression of two functions whose limit cannot be evaluated by direct substitution.
  • The most common indeterminate forms are \(0/0\) and \( \pm\infty/\pm\infty\).
  • The other indeterminate forms refer to the expressions \(0 \cdot \infty\), \(0^0\), \( \infty^0\), \(1^\infty\), and \(\infty-\infty\).
• A limit resulting in an indeterminate form that involves quotients can be evaluated using L'Hôpital's Rule.
• By algebraic means, it is possible to transform limits that involve the other indeterminate forms into expressions that involve quotients, so you can use L'Hôpital's rule to evaluate them.

\[ \lim_{x \to 0^+} \left( \frac{1}{x}-\frac{1}{x^2} \right) = -\infty\]

Evaluate the limit

\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right)\]

Solution:

Once again, if you were to evaluate the limit directly, you would find that:

\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) = \infty-\infty\]

By subtracting the fractions, you get:

\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) = \lim_{x \to 0^+} \frac{\cos{x}-1}{x}\]

The cosine of \(0\) is \(1,\) so both the numerator and the denominator approach \(0\) as \(x \to 0.\) This suggests the use of L'Hôpital's rule, that is:

\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) = \lim_{x \to 0^+}\frac{\sin{x}}{1}\]

Since the sine of \(0\) is \(0\), you can now evaluate the limit, obtaining:

\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) =0\]

Evaluate the limit

\[ \lim_{x \to \infty} x\, e^{-x}.\]

Solution:

As usual, begin by trying to evaluate the limit directly. The limit as \(x \to \infty\) of \(e^{-x}\) is \(0\), so you are dealing with an indeterminate form of \( \infty \cdot 0\). To use L'Hôpital, note that you can write \(e^{-x}\) as \(e^x\) in the denominator, that is

\[ \lim_{x \to \infty} x\,e^{-x} = \lim_{x \to \infty}\frac{x}{e^x}.\]

Now you have an indeterminate form of \( \infty/\infty\), so use L'Hôpital's rule,

\[ \begin{align} \lim_{x \to \infty} x\,e^{-x} &= \lim_{x \to \infty} \frac{1}{e^x}\ \\ &= 0. \end{align}\]

Evaluate the limit

\[\lim_{x \to 0^+} x^x.\]

Solution:

Begin by labeling the limit, namely

\[ L = \lim_{x \to 0^+}x^x.\]

From here, you can take the natural logarithm of both sides, that is

\[ \ln{L}=\ln{\left( \lim_{x \to 0^+} x^x \right)}.\]

Because the natural logarithm is a continuous function, you can move it inside the limit and use the properties of natural logarithms, so

\[ \begin{align} \ln{L} &= \lim_{x \to 0^+} \left( \ln{x^x} \right) \\ &= \lim_{x \to 0^+} x\ln{x}. \end{align}\]

As \(x \to 0^+\), the natural logarithm goes to negative infinity, so the above expression is an indeterminate form of \(0 \cdot \infty\), which you can work using some algebra

\[ \begin{align} \ln{L} &= \lim_{x \to 0^+} x\ln{x} \\ &= \lim_{x \to 0^+} \frac{\ln{x}}{\frac{1}{x}}, \end{align}\]

and then use L'Hôpital's rule, so

\[ \begin{align} \ln{L} &= \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} \\ &= \lim_{x \to 0^+} (-x) \\ &= 0. \end{align} \]

Finally, undo the natural logarithm by using the exponential function, so

\[ \begin{align} L &= e^0 \\ &= 1. \end{align}\]

Evaluate the limit

\[\lim_{x \to 0^+} \left(\frac{1}{x}-\csc{x} \right).\]

Solution:

Begin by recalling that the cosecant function is the reciprocal of the sine function, so

\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\csc{x} \right) = \lim_{x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right).\]

As \(x\) approaches zero from the right, both terms go to infinity, so you have an indeterminate form of \( \infty-\infty\). Work this around by subtracting the fractions

\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) = \lim_{x \to 0^+} \left( \frac{\sin{x}-x}{x\sin{x}}\right),\]

which is now an indeterminate form of \(0/0\). Use L'Hôpital's rule, that is

\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) = \lim_{x \to 0^+} \frac{\cos{x}-1}{\sin{x}+x\cos{x}},\]

but you will find that this is another indeterminate form of \(0/0\). Use L'Hôpital's rule once more, so

\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) = \lim_{x \to 0^+} \frac{\sin{x}}{\cos{x}+\cos{x}-x\sin{x}},\]

which can now be evaluated, giving you

\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) =0.\]

Evaluate the limit

\[ \lim_{x \to 0^+} x\cot{x}.\]

Solution:

Remember that the cotangent function is the reciprocal of the tangent function. Since \(\tan{0}=0\), the cotangent goes to infinity when approached from the right, so this is an indeterminate form of \(0 \cdot \infty.\) To solve this, rewrite the cotangent as the reciprocal of the tangent, that is

\[ \lim_{x \to 0^+} x\cot{x} = \lim_{x \to 0^+} \frac{x}{\tan{x}},\]

which is now an indeterminate form of \(0/0\), so use L'Hôpitals rule

\[ \lim_{x \to 0^+} x\cot{x} = \lim_{x \to 0^+} \frac{1}{\sec^2{x}}.\]

The secant of \(0\) is equal to \(1\), so

\[ \lim_{x \to 0^+} x\cot{x}=1.\]

Evaluate the limit

\[ \lim_{x \to \infty}x^{^1/_x}. \]

Solution:

As \(x\) goes to infinity, \(1/x\) goes to zero, so this is an indeterminate form of \(\infty^0\). Label the limit as \(L\) and find its natural logarithm, that is

\[ \ln{L} = \ln{\left( \lim_{x \to \infty} x^{^1/_x} \right)}, \]

and use the fact that the natural logarithm is a continuous function to introduce it inside the limit, so

\[ \ln{L} = \lim_{ x\to \infty} \ln{\left( x^{^1/_x}\right)}.\]

Now, use the properties of logarithms to write

\[ \begin{align} \ln{L} &= \lim_{x \to \infty} \left( \frac{1}{x} \ln{x}\right) \\ &= \lim_{x \to \infty} \frac{\ln{x}}{x}\end{align}.\]

The above limit is now an indeterminate form of \(\infty/\infty\), so you can use L'Hôpital's rule, obtaining

\[ \begin{align} \ln{L} &= \lim_{x \to \infty} \frac{\frac{1}{x}}{1} \\ &=\frac{0}{1} \\&= 0.\end{align}\]

Finally, undo the natural logarithm by taking the exponential, which means that

\[ \begin{align} L &= e^0 \\ &= 1. \end{align} \]