Question

A doubt in Limits.......​

Answer 1

$\sf \longrightarrow \lim \limits_{ \sf x \rightarrow1}{ \sf \dfrac{x + {x}^{2} + {x}^{3} + {x}^{4} - 4}{x - 1} } \\ \\ \\ \sf \longrightarrow \lim \limits_{ \sf x \rightarrow1}{ \sf \dfrac{(x - 1) + ({x}^{2} - 1)+ ({x}^{3} - 1)+ ({x}^{4} - 1)}{x - 1} }\\ \\ \\ \sf \longrightarrow \lim \limits_{ \sf x \rightarrow1}{ \sf \dfrac{(x - 1) + (x - 1) ({x} + 1)+ ( x - 1)({x}^{2} + x + 1)+({x} + 1) ({x} - 1) ({x}^{2} + 1)}{x - 1} }\\ \\ \\ \sf \longrightarrow \lim \limits_{ \sf x \rightarrow1}{ \sf \dfrac{(x - 1) \bigg(1+ ({x} + 1)+ ({x}^{2} + x + 1)+({x} + 1) ({x}^{2} + 1)\bigg)}{x - 1} }\\ \\ \\ \sf \longrightarrow \lim \limits_{ \sf x \rightarrow1}{ \sf \dfrac{ 1+ ({x} + 1)+ ({x}^{2} + x + 1)+({x} + 1) ({x}^{2} + 1)}{1} }\\ \\ \\ \sf \longrightarrow { \sf \dfrac{ 1+ 1+ 1+ {1}^{2} + 1+ 1+({1} + 1) ({1}^{2} + 1)}{1} }\\ \\ \\ \sf \longrightarrow { \sf \dfrac{ 1+ 1+ 1+ 1 + 1+ 1+(2 \times 2 )}{1} }\\ \\ \\ \sf \longrightarrow { \sf \dfrac{ 10}{1} } = 10$

Answer : 10

Answer 2

Answer:

That's it my friend.

BTW,from same class.........:)