Question

$\underset{x\to \infty }{lim}(e^{11x}-7x{)}^{\frac{1}{3x}}$ is equal to:

• A. $y=e^{\frac{1}{3}}$
• B. $y=-e^{\frac{11}{3}}$
• C. $y=e^{\frac{11}{3}}$
• D. None of these

Answer 1

Answer: (C)

$y=e^{\frac{11}{3}}$

$\underset{n\to \omega }{lim}(e^{11x}-7x{)}^{\frac{1}{3x}}$

$lety=lim{\left(e^{11x}-7x\right)}^{\frac{1}{3x}}$

$\log y=lim\frac{1}{3x}\log \left(e^{11x}-7x\right)$

= $\underset{n\to \infty }{lim}\frac{\log \left(e^{11x}-7x\right)}{3x}$ $\left(\omega /\infty \right)$

= $\frac{\frac{\left(11e^{11x}-7\right)}{\left(e^{11x}-7x\right)}}{3}$

= $\frac{1}{3}\frac{\left(11e^{11x}-7\right)}{e^{11x}-7x}$ $\left(\omega /\infty \right)$

= $\frac{11}{3}.\frac{11e^{11x}}{11e^{11x}-7}$ $\left(\omega /\infty \right)$

= $\frac{121}{3}\times \frac{11e^{11x}}{121e^{11x}}$

= $\frac{11}{3}$

$y=e^{\frac{11}{3}}$