How to calculate this? $lim_(x->0)(ln(1+7x^2)/(7x^2)-ln(1+2x)/(2x))$ ; $x>0$ .

Question

999 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-10T17:21:02

$0$

First, we know that $log_e(1) = 0$ and also

$lim_(x->0)(log_e(1+x)-log_e(1))/x = d/(dx)log x]_(x = 1) = 1$

so

$lim_(x->0)(ln(1+7x^2)/(7x^2)-ln(1+2x)/(2x)) = d/(dx)log_e(1) -d/(dx)log_e(1)=1-1=0$

How to calculate this? lim_(x->0)(ln(1+7x^2)/(7x^2)-ln(1+2x)/(2x))lim_(x->0)(ln(1+7x^2)/(7x^2)-ln(1+2x)/(2x));x>0x>0.