Question

lim x tend ti zero 1 + x 1/x = e​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{\lim_{x\to0}\,\left(1+x\right)^{\frac{1}{x}}}$

$\displaystyle\tt{=e^{\displaystyle\lim_{x\to0}\,\log\left\{\left(1+x\right)^{\frac{1}{x}}\right\}}}$

$\displaystyle\tt{=e^{\displaystyle\lim_{x\to0}\,\dfrac{1}{x}\cdot\log\left(1+x\right)}}$

$\displaystyle\tt{=e^{\displaystyle\lim_{x\to0}\,\dfrac{\log\left(1+x\right)}{x}}}$

$\displaystyle\tt{=e^{1}}$

$\tt{=e}$