The function f:R/{0}→R given by f(x)=1x−2e2x−1 can be made continuous at x=0  by defining f(0)  as

Given f(x) is continuous at x=0⇒f(0)=Ltx→01x−2e2x−1=Ltx→0(e2x−1)−2xx(e2x−1) Apply L Hospital Rule=Ltx→02e2x−2e2x−1+x⋅2e2x 00 form by L'Hospital rule=Ltx→04e2x2e2x+2[1⋅e2x+2xe2x]=42+2=1 .