Question

Evaluate : $\underset{x\to 0}{lim}\frac{2^x{.3}^x{.5}^x-7^x{.11}^x}{{11}^x{.5}^x-2^x{.3}^x{.7}^x}$

• A. ${\log }_{30/77}\left(\frac{55}{45}\right)$
• B. ${\log }_{55/42}\left(\frac{30}{77}\right)$
• C. ${\log }_{6/7}\left(\frac{5}{7}\right)$
• D. ${\log }_{5/7}\left(\frac{6}{7}\right)$

Answer 1

The correct option is B ${\log }_{55/42}\left(\frac{30}{77}\right)$

$\underset{x\to 0}{lim}\frac{2^x{.3}^x{.5}^x-7^x{.11}^x}{{11}^x{.5}^x-2^x{.3}^x{.7}^x}$

$=\underset{{}_{x\to 0}}{lim}\frac{(30{)}^x-(77{)}^x}{(55{)}^x-(42{)}^x}$

$=\underset{{}_{x\to 0}}{lim}\frac{(30{)}^x-1+1-(77{)}^x}{(55{)}^x-1+1-(42{)}^x}$

$=\underset{{}_{x\to 0}}{lim}\frac{\frac{(30{)}^x-1}{x}-\left(\frac{(77{)}^x-1}{x}\right)}{\frac{(55{)}^x-1}{x}-\left(\frac{(42{)}^x-1}{x}\right)}$

$=\frac{\log 30-\log 77}{\log 55-\log 42}$ $[byL.Hospitalrule]$

$=\frac{\log \frac{30}{77}}{\log \frac{55}{42}}={\log }_{\left(55/42\right)}\left(\frac{30}{77}\right)$