Q. 37 - Calculus

Question

Calculate each of the limits in Exercises $21-48$. Some of these limits are made easier by L’Hopital’s rule, and some are not.

$\underset{x\to 0^{+}}{\lim }\frac{{\log }_{2}x}{{\log }_{2}3x}$.

Step-by-Step Solution

Answer

The value of the limit $\underset{x\to 0^{+}}{\lim }\frac{{\log }_{2}x}{{\log }_{2}3x}$ is, $1$.

Step 1 . Given information

$\underset{x\to 0^{+}}{\lim }\frac{{\log }_{2}x}{{\log }_{2}3x}$.

Step 2 . The limit is given below:

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\frac{{\log }_{2}x}{{\log }_{2}3x}=\underset{x\to 0^{+}}{\lim }\frac{{\log }_{2}x}{{\log }_{2}3+{\log }_{2}x} \\ =\underset{x\to 0^{+}}{\lim }\frac{{\log }_{2}x}{{\log }_{2}x\left(\frac{{\log }_{2}3}{{\log }_{2}x}+1\right)} \\ =\underset{x\to 0^{+}}{\lim }\frac{1}{\left(\frac{{\log }_{2}3}{{\log }_{2}x}+1\right)} \\ =\underset{x\to 0^{+}}{\lim }\frac{1}{(0+1)} \\ =1 \end{gathered}$$

The exact value of the limit is, $1$.