If-lim-x-c-a-0-a-1-x-c-a-2-x-c-2-a-n-x-c-n-x-c-n-0-then-a-0-a-1-a-2-a-n-

Question Number 55634 by gunawan last updated on 01/Mar/19

If \lim_{x \to c} \frac{a_{0} + a_{1} (x - c) + a_{2} {(x - c)}^{2} + \dots + a_{n} {(x - c)}^{n}}{{(x - c)}^{n}} = 0

then a_{0} + a_{1} + a_{2} + . . + a_{n} = . .

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19

t = x - c

\lim_{t \to 0} \frac{a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + \dots + a_{n} t^{n}}{t^{n}}

f o r (\frac{0}{0}) f o r m a_{0} = 0

\lim_{t \to 0} \frac{a_{1} + 2 a_{2} t + 3 a_{3} t^{2} + \dots + {n a}_{n} t^{n - 1}}{{n t}^{n - 1}}

t o m a k e (\frac{0}{0}) f o r m a_{1} \to 0

s o i n m y v i e w \underset{i = 0}{\sum^{n}} a_{i} = 0

l e t o t h e r s c h e c k \dots

Commented by gunawan last updated on 01/Mar/19

yes

thank you Sir

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19

m o s t w e l c o m e \dots p l s c h e c k t h e o t h e r s s o l v e d p r o b l e m o r

u p l o a d t h e a n s w e r o n l y \dots n o t d e t a i l s \dots