Question

$\underset{x\to 0}{lim}\frac{{10}^x-2^x-5^x+1}{xtanx}$ is equal to -

• A. $\ln (2+5)$
• B. $\ln 5\ln 2$
• C. $\ln 10$
• D. $\ln \left(\begin{array}{c}\frac{5}{2}\end{array}\right)$

Answer 1

Answer: (B)

$\ln 5\ln 2$

By L-Hospital's Rule,

$=\frac{{10}^{x}ln10-2^{x}ln2-5^{x}ln5}{xse{c}^{2}x+tanx}=0/0form$

$=\frac{{10}^{x}ln10.ln10-2^{x}ln2.ln2-5^{x}ln5.ln5}{2xse{c}^{2}xtanx+se{c}^{2}x+se{c}^2}$

$=\frac{ln\left(5.2\right).ln\left(5.2\right)-ln2.ln2-ln5.ln5}{2}$

$=\frac{ln5.ln5+ln2.ln2+2ln5ln2-ln2.ln2-ln5.ln5}{2}$

$=ln5.ln2$