lim[(x+6)/(x+1)] ^x+4 is equal to

x tends to infinity

we will use the concept of 1 : f(x)^g(x) = e ^{lim x tends(f(x) - 1)(g(x))}

so in the question e^{lim x tends to [( x + 6)/ (x+1) - 1] (x + 4)}

e^{lim x tends to [x+6 - x - 1 /x+1] x+4}

e^{lim x tends to [5/x+1] ( x+4)}

e^{lim x tends to 5x+20 / x+1}

taking x common

e^{lim x tends to [x(5+20/x) / x(1+1/x)]}

x get cancelled and by placing in place of x we will get 20/x and 1/x= 0

e⁵ is the answer.