Question Number 131882 by Eric002 last updated on 09/Feb/21

Answered by liberty last updated on 09/Feb/21
![L′Ho^ pital L= lim_(x→0) [ (1/(1+(1/(4 (((x+1)^3 ))^(1/4) )))) ]=(1/(1+(1/4)))= (4/5) or lim_(x→0) [ (x/(x+1+(x/4)−1))] = lim_(x→0) ((4x)/(5x))=(4/5)](https://www.tinkutara.com/question/Q131886.png)
Commented by Eric002 last updated on 09/Feb/21

Commented by EDWIN88 last updated on 09/Feb/21

Commented by Eric002 last updated on 09/Feb/21

Answered by EDWIN88 last updated on 09/Feb/21

Answered by mathmax by abdo last updated on 10/Feb/21
