Question

lim u tends to ∞[e^2u]/[e^u + u^2]

Answer 1

\[\infty\]

Answer 2

we can use l'hospital (however you spell it)

\[\lim_{u \rightarrow \infty} \frac{2e^{2u}}{e^u+2u}=\lim_{u \rightarrow \infty} \frac{4e^{2u}}{e^u+2}=\lim_{u \rightarrow \infty}\frac{8e^{2u}}{e^u}=\lim_{u \rightarrow \infty}8e^{u}=\infty\]

Answer 3

i hope u see i use law of exponents
8e^{2u}/e^u=8e^{2u-u}=8e^u

Answer 4

ok.great.