Evaluate the following limits:

a$.\underset{x{0}^{+}}{lim}xln(x)=$

b$.\underset{x{0}^{+}}{lim}{x}^x=$

c$.\underset{x}{lim}\frac{ln(5x)}{5x}=$

d$.\underset{x25}{lim}\frac{x-25}{\sqrt[\phantom{2}]{x}-5}=$

e$.\underset{x-}{lim}{x}^2{e}^x=$

f$.\underset{x}{lim}\frac{x^2+x-1}{\sqrt[\phantom{2}]{4x^4+5x+1}}=($what is the HA$?)$