Chapter 9.2, Problem 36E

To determine

To Find: The value of a limit using L'Hôpital's Rule.

Answer to Problem 36E

The value of the limit is $1$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{y\to 0^{+}}{\lim }\frac{\ln \left(y^2+2y\right)}{\ln y}$ .

Method: $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to a,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: As $y\to 0^{+}$ ,

  $\ln \left(y^2+2y\right)\to -\infty$ and $\ln y\to -\infty$

Therefore, the limit is of the indeterminate form $\frac{\infty }{\infty }$

So, apply L'Hôpital's Rule to write

  $\begin{array}{l}\underset{y\to 0^{+}}{\lim }\frac{\ln \left(y^2+2y\right)}{\ln y}=\underset{y\to 0^{+}}{\lim }\frac{\left(\raisebox{1ex}{$2y+2$}\!\left/ \!\raisebox{-1ex}{$y^2+2y$}\right.\right)}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$y$}\right.\right)}=\underset{y\to 0^{+}}{\lim }\left(\frac{2y+2}{y+2}\right)\\ =\frac{2\cdot 0+2}{0+2}=\frac{2}{2}=1\end{array}$

Therefore, the value of the limit is $1$ .