If the function f defined as f(x) = 1/(x - k√e^(2x)), x ≠ 0, is continuous at x = 0, then the ordered pair (k, f(0)) is equal to?

If the function is continuous atx=0,limx→0f(x)exists and is equal tof(0)limx→0f(x)=limx→01x−k𕒵e2x𕒵=limx→0e2x𕒵−kx+x(x)(e2x𕒵)Using Taylor's expansion fore2x, we get=limx→0(1+2x+(2x)22!+(2x)33!+..)𕒵−kx+x(x)((1+2x+(2x)22!+(2x)33!+..)𕒵)=limx→0(3−k)x+4x22!+8x33!+.. (2x2+4x32!+8x43!+..)For the limit to exist, power ofxin the numerator should be greator than or equal to the power ofxin the denominator.Therefore, coefficient ofxin numerator is equal to zero⟹3−k=0⟹k=3So the limit reduces tolimx→0(x2)(42!+8x3!+..)(x2)(2+4x2!+8x23!+..)=limx→042!+8x3!+...2+4x2!+8x23!+...=1Hence,f(0)=1Therefore, answer is option (A).