Giúp mình với L=lim(2n^4+1)^2(n+2)^9/n^17+1 28/07/2021 Bởi Mackenzie Giúp mình với L=lim(2n^4+1)^2(n+2)^9/n^17+1
Giải thích các bước giải: $L=lim\dfrac{(2n^4+1)^2.(n+2)^9}{n^{17}+1}$ $\rightarrow L=lim\dfrac{\dfrac{(2n^4+1)^2}{n^8}.\dfrac{(n+2)^9}{n^9}}{1+\dfrac{1}{n^{17}}}$ $\rightarrow L=lim\dfrac{(\dfrac{2n^4+1}{n^4})^2.(\dfrac{n+2}{n})^9}{1+\dfrac{1}{n^{17}}}$ $\rightarrow L=lim\dfrac{(2+\dfrac{1}{n^4})^2.(1+\dfrac{2}{n})^9}{1+\dfrac{1}{n^{17}}}$ $\rightarrow L=\dfrac{(2+0)^2.(1+0)^9}{1+0}$ $\rightarrow L=4$ Bình luận
Giải thích các bước giải:
$L=lim\dfrac{(2n^4+1)^2.(n+2)^9}{n^{17}+1}$
$\rightarrow L=lim\dfrac{\dfrac{(2n^4+1)^2}{n^8}.\dfrac{(n+2)^9}{n^9}}{1+\dfrac{1}{n^{17}}}$
$\rightarrow L=lim\dfrac{(\dfrac{2n^4+1}{n^4})^2.(\dfrac{n+2}{n})^9}{1+\dfrac{1}{n^{17}}}$
$\rightarrow L=lim\dfrac{(2+\dfrac{1}{n^4})^2.(1+\dfrac{2}{n})^9}{1+\dfrac{1}{n^{17}}}$
$\rightarrow L=\dfrac{(2+0)^2.(1+0)^9}{1+0}$
$\rightarrow L=4$