If `f(x) = {{:(sin x"," x ne npi"," n = 0"," pm1"," pm2","...),(" 2, ""otherwise"):}}` `and g (x) ={{:(x^(2)+1"," x ne 0","2),(" 4, "x=0),(" 5, "x=2):}},"then" lim_(x to 0) g[f(x)]` is ………

If `f(x) = {{:(sin x"," x ne npi"," n = 0"," pm1"," pm2","...),(" 2, "

1.	If `f(x) = {{:(sin x"," x ne npi"," n = 0"," pm1"," pm2","...),(" 2, ""otherwise"):}}` `and g (x) ={{:(x^(2)+1"," x ne 0","2),(" 4, "x=0),(" 5, "x=2):}},"then" lim_(x to 0) g[f(x)]` is ………
Answer» Correct Answer - 1 Given, `f(x) ={{:(sin x", "x ne npi", n=0,"pm"1," pm"2,"...),(" 2, otherwise"):}` `g[f(x)]={{:({f(x)}^(2)+1" , "f(x) ne 0","2),(" 4 , "f(x) = 0),(" 5 , "f(x) = 2):}` `:. g[f(x)]={{:((sin^(2) x) +1"," x ne"," n pi = 0"," pm 1","...),(" 5,"x=npi):}` Now, `underset( x to 0) lim [f(x)] = underset( x to 0) lim (sin^(2)x) + 1 = 1`