Answer
\[e^4\]
Work Step by Step
Let \[l=\lim_{x\rightarrow \infty}\left(1+\frac{4}{x}\right)^x\]
\[\Rightarrow \ln l=\lim_{x\rightarrow \infty}x\ln \left(1+\frac{4}{x}\right)\]
\[\Rightarrow \ln l=\lim_{x\rightarrow \infty}\frac{\ln \left(1+\frac{4}{x}\right)}{\frac{1}{x}}\]
Which is $\frac{0}{0}$ form
Using L' Hopitals rule
\[\Rightarrow \ln l=\lim_{x\rightarrow \infty}\frac{\{\ln \left(1+\frac{4}{x}\right)\}'}{(\frac{1}{x})'}\]
\[\Rightarrow \ln l=\lim_{x\rightarrow \infty}\frac{\frac{\frac{-4}{x^2}}{1+\frac{4}{x}}}{{\frac{-1}{x^2}}}\]
\[\Rightarrow \ln l=\lim_{x\rightarrow \infty}\frac{4}{1+\frac{4}{x}}=4\]
\[\Rightarrow l=e^4\]
Hence , \[\lim_{x\rightarrow \infty}\left(1+\frac{4}{x}\right)^x=e^4\]
