4.8 L’hôpital’s rule  (Page 4/7)

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

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

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

Therefore,

Let $y=x^{1\text{/}x}.$ Then,

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

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

which leads to

Hence,

Let

Therefore,

We now evaluate $\underset{x\to 0^{+}}{\text{lim}}\text{sin}x\text{ln}x.$ Since $\underset{x\to 0^{+}}{\text{lim}}\text{sin}x=0$ and $\underset{x\to 0^{+}}{\text{lim}}\text{ln}x=\text{−}\infty ,$ we have the indeterminate form $0·\infty .$ To apply L’Hôpital’s rule, we need to rewrite $\text{sin}x\text{ln}x$ 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 $0·\infty ,$ 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 $\text{csc}x=\frac{1}{\text{sin}x}$ and $\text{cot}x=\frac{\text{cos}x}{\text{sin}x},$ we can rewrite the expression on the right-hand side as

We conclude that $\underset{x\to 0^{+}}{\text{lim}}\text{ln}y=0.$ Therefore, $\text{ln}\left(\underset{x\to 0^{+}}{\text{lim}}y\right)=0$ and we have

Hence,