1. Since the numerator [latex]1- \cos x\to 0[/latex] and the denominator [latex]x\to 0,[/latex] we can apply L’Hôpital’s rule to evaluate this limit. We have
   [latex]\begin{array}{cc}\hfill \underset{x\to 0}{\text{lim}}\frac{1- \cos x}{x}& =\underset{x\to 0}{\text{lim}}\frac{\frac{d}{dx}(1- \cos x)}{\frac{d}{dx}(x)}\hfill \\ & =\underset{x\to 0}{\text{lim}}\frac{ \sin x}{1}\hfill \\ & =\frac{\underset{x\to 0}{\text{lim}}( \sin x)}{\underset{x\to 0}{\text{lim}}(1)}\hfill \\ & =\frac{0}{1}=0.\hfill \end{array}[/latex]
2. As [latex]x\to 1,[/latex] the numerator [latex]\sin (\pi x)\to 0[/latex] and the denominator [latex]\text{ln}(x)\to 0.[/latex] Therefore, we can apply L’Hôpital’s rule. We obtain
   [latex]\begin{array}{cc}\hfill \underset{x\to 1}{\text{lim}}\frac{ \sin (\pi x)}{\text{ln}x}& =\underset{x\to 1}{\text{lim}}\frac{\pi \cos (\pi x)}{1\text{/}x}\hfill \\ & =\underset{x\to 1}{\text{lim}}(\pi x) \cos (\pi x)\hfill \\ & =(\pi ·1)(-1)=\text{−}\pi .\hfill \end{array}[/latex]
3. As [latex]x\to \infty ,[/latex] the numerator [latex]{e}^{1\text{/}x}-1\to 0[/latex] and the denominator [latex](\frac{1}{x})\to 0.[/latex] Therefore, we can apply L’Hôpital’s rule. We obtain
   [latex]\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\frac{1}{x}}=\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}(\frac{-1}{{x}^{2}})}{(\frac{-1}{{x}^{2}})}=\underset{x\to \infty }{\text{lim}}{e}^{1\text{/}x}={e}^{0}=1·\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\text{ln}x}.[/latex]
4. As [latex]x\to 0,[/latex] both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
   [latex]\underset{x\to 0}{\text{lim}}\frac{ \sin x-x}{{x}^{2}}=\underset{x\to 0}{\text{lim}}\frac{ \cos x-1}{2x}.[/latex]

   Since the numerator and denominator of this new quotient both approach zero as [latex]x\to 0,[/latex] we apply L’Hôpital’s rule again. In doing so, we see that

   [latex]\underset{x\to 0}{\text{lim}}\frac{ \cos x-1}{2x}=\underset{x\to 0}{\text{lim}}\frac{\text{−} \sin x}{2}=0.[/latex]

   Therefore, we conclude that

   [latex]\underset{x\to 0}{\text{lim}}\frac{ \sin x-x}{{x}^{2}}=0.[/latex]

1. Since [latex]3x+5[/latex] and [latex]2x+1[/latex] are first-degree polynomials with positive leading coefficients, [latex]\underset{x\to \infty }{\text{lim}}(3x+5)=\infty[/latex] and [latex]\underset{x\to \infty }{\text{lim}}(2x+1)=\infty .[/latex] Therefore, we apply L’Hôpital’s rule and obtain
   [latex]\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3}{2}=\frac{3}{2}.[/latex]

   Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[/latex] in the denominator. In doing so, we saw that

   [latex]\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2x+1}=\frac{3}{2}.[/latex]

   L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.

2. Here, [latex]\underset{x\to {0}^{+}}{\text{lim}}\text{ln}x=\text{−}\infty[/latex] and [latex]\underset{x\to {0}^{+}}{\text{lim}} \cot x=\infty .[/latex] Therefore, we can apply L’Hôpital’s rule and obtain
   [latex]\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}x}{ \cot x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{\text{−}{ \csc }^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{−}x{ \csc }^{2}x}.[/latex]

   Now as [latex]x\to {0}^{+},[/latex] [latex]{ \csc }^{2}x\to \infty .[/latex] Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\csc x[/latex] to write

   [latex]\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{−}x{ \csc }^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{{ \sin }^{2}x}{\text{−}x}.[/latex]

   Now [latex]\underset{x\to {0}^{+}}{\text{lim}}{ \sin }^{2}x=0[/latex] and [latex]\underset{x\to {0}^{+}}{\text{lim}}x=0,[/latex] so we apply L’Hôpital’s rule again. We find

   [latex]\underset{x\to {0}^{+}}{\text{lim}}\frac{{ \sin }^{2}x}{\text{−}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{2 \sin x \cos x}{-1}=\frac{0}{-1}=0.[/latex]

   We conclude that

   [latex]\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}x}{ \cot x}=0.[/latex]

Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get

and

At which point we would conclude erroneously that

However, since [latex]\underset{x\to 1}{\text{lim}}({x}^{2}+5)=6[/latex] and [latex]\underset{x\to 1}{\text{lim}}(3x+4)=7,[/latex] we actually have

We can conclude that

First, rewrite the function [latex]x\text{ln}x[/latex] as a quotient to apply L’Hôpital’s rule. If we write

we see that [latex]\text{ln}x\to \text{−}\infty[/latex] as [latex]x\to {0}^{+}[/latex] and [latex]\frac{1}{x}\to \infty[/latex] as [latex]x\to {0}^{+}.[/latex] Therefore, we can apply L’Hôpital’s rule and obtain

We conclude that

Figure 2. Finding the limit at [latex]x=0[/latex] of the function [latex]f(x)=x\text{ln}x.[/latex]

By combining the fractions, we can write the function as a quotient. Since the least common denominator is [latex]{x}^{2} \tan x,[/latex] we have

As [latex]x\to {0}^{+},[/latex] the numerator [latex]\tan x-{x}^{2}\to 0[/latex] and the denominator [latex]{x}^{2} \tan x\to 0.[/latex] Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

As [latex]x\to {0}^{+},[/latex] [latex]({ \sec }^{2}x)-2x\to 1[/latex] and [latex]{x}^{2}{ \sec }^{2}x+2x \tan x\to 0.[/latex] Since the denominator is positive as [latex]x[/latex] approaches zero from the right, we conclude that

Therefore,

Let [latex]y={x}^{1\text{/}x}.[/latex] Then,

We need to evaluate [latex]\underset{x\to \infty }{\text{lim}}\frac{\text{ln}x}{x}.[/latex] Applying L’Hôpital’s rule, we obtain

Therefore, [latex]\underset{x\to \infty }{\text{lim}}\text{ln}y=0.[/latex] Since the natural logarithm function is continuous, we conclude that

which leads to

Hence,

Let

Therefore,

We now evaluate [latex]\underset{x\to {0}^{+}}{\text{lim}} \sin x\text{ln}x.[/latex] Since [latex]\underset{x\to {0}^{+}}{\text{lim}} \sin x=0[/latex] and [latex]\underset{x\to {0}^{+}}{\text{lim}}\text{ln}x=\text{−}\infty ,[/latex] we have the indeterminate form [latex]0·\infty .[/latex] To apply L’Hôpital’s rule, we need to rewrite [latex]\sin x\text{ln}x[/latex] as a fraction. We could write

or

Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain

Unfortunately, we not only have another expression involving the indeterminate form [latex]0·\infty ,[/latex] but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing

and applying L’Hôpital’s rule, we obtain

Using the fact that [latex]\csc x=\frac{1}{ \sin x}[/latex] and [latex]\cot x=\frac{ \cos x}{ \sin x},[/latex] we can rewrite the expression on the right-hand side as

We conclude that [latex]\underset{x\to {0}^{+}}{\text{lim}}\text{ln}y=0.[/latex] Therefore, [latex]\text{ln}(\underset{x\to {0}^{+}}{\text{lim}}y)=0[/latex] and we have

Hence,