f(0)=x→0limtanx−xetanx−ex+ln(secx+tanx)−x
Applying L'hospital rule
⇒f(0)=x→0limsec2x−1etanx⋅sec2x−ex+secx−1
⇒f(0)=x→0limtan2xetanx(sec2x−1)+(etanx−ex)+secx−1
⇒f(0)=x→0lim(etanx+tan2xex(etanx−x−1)+secx+11)
⇒f(0)=1+0+21=23
JEE Main 2026 — Mathematics Calculus
If the function f(x)=tanx−xex(etanx−x−1)+loge(secx+tanx)−x is continuous at x=0, then the value of f(0) is equal to
Held on 24 Jan 2026 · Verified 6 Jul 2026.
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