Question

$\underset{x\to 0^{+}}{lim}\left\{In\right(x^7+2x^{6}e+x^5{e}^2)-In(x^2{\sin }^{3}x\left)\right\}$ is equal to

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

$\underset{x\to 0^{+}}{lim}$ $\{ln(x^7+{2x}^{6}e+x^5{e}^2)-ln\left(x^2{\sin }^{3}x\right)\}$

$=\underset{x\to 0^{+}}{lim}$ $\left\{ln\left(\frac{x^7+{2x}^{6}e+x^5{e}^2}{x^2{\sin }^{3}x}\right)\right\}$

$=\underset{x\to 0^{+}}{lim}$ $\left\{ln\left(\frac{x^5+{2x}^{4}e+x^3{e}^2}{{\sin }^{3}x}\right)\right\}$

$=\underset{x\to 0^{+}}{lim}$ $\left\{ln\left(\frac{x^3(x^2+2xe+e^2)}{{\sin }^{3}x}\right)\right\}$

$=\underset{x\to 0^{+}}{lim}$ $\left\{ln\left(\frac{x^2+2xe+e^2}{\frac{{\sin }^{3}x}{x^3}}\right)\right\}$

$=ln\left\{\frac{\underset{x\to 0^{+}}{lim}{x}^2+2xe+e^2}{\underset{x\to 0^{+}}{lim}{\left(\frac{sinx}{x}\right)}^3}\right\}$

$=ln{e}^2$ $(\because \underset{x\to 0}{lim}{\left(\frac{sinx}{x}\right)}^3=1)$

$=ln{e}^2$

$=2lne$

$=2$ $(\because lne=1)$