Ts Eamcet
Limits Continuity And Differentiability
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ+1 / -02024
Define $ f: R \rightarrow R $ by $ f(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^{2}}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0\end{array}\right. $
Then, the value of $ a $ so that $ f $ is continuous at $ x=0 $ is
