0 votes 0 votes $\underset{x\rightarrow 1}{\lim} \dfrac{x^{16}-1}{\mid x-1\mid}$ equals $-1$ $0$ $1$ Does not exist Calculus isi2016-dcg calculus limits + – gatecse asked Sep 18, 2019 • recategorized Nov 19, 2019 by Lakshman Bhaiya gatecse 547 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Verma Ashish commented Oct 12, 2019 i edited by Verma Ashish Oct 12, 2019 reply Follow Share Ohhh.. i made a mistake $LHL=\underset{h\rightarrow 0}{\lim} \dfrac{(1-h)^{16}-1}{h}$= -16 $RHL= 16$ So LHL$\neq$RHL Waht will be value of function at x=1, will it be not defined?? 2 votes 2 votes ankitgupta.1729 commented Oct 12, 2019 reply Follow Share yes, it is undefined at x=1 1 votes 1 votes techbd123 commented Oct 12, 2019 reply Follow Share @Verma Ashish Yeah. The limit doesn't exist because LHL $\ne$ RHL. It can be identified by this below. $x \to 1^{-}\Rightarrow |x-1|=-(x-1)$ whereas $x \to 1^{+}\Rightarrow |x-1|=(x-1)$ 0 votes 0 votes Please log in or register to add a comment.
