Finding a tricky limit

Solution 1:

Note that if $$f(x)=\int_{0}^{x}(1+\sin t) ^{1/t}\,dt=\int_{0}^{x}g(t)\,dt$$ then the limit in question is $f'(0)$. By fundamental theorem of calculus we have $f'(0)=g(0)$ provided $g$ is continuous at $0$. Luckily the integrand has a removable discontinuity at $0$ and by defining $$g(0)=\lim_{t\to 0}(1+\sin t) ^{1/t}=e$$ we can ensure that $g$ is continuous at $0$ and therefore $f'(0)=g(0)=e$.

Usage of L'Hospital's Rule is totally unnecessary here and I wonder why L'Hospital's Rule was given as a hint. Also the fact that the integrand $(1+\sin t) ^{1/t}$ has a removable discontinuity at $0$ is important to get an answer to this question, but it is not needed for "interpreting the given integral as a Riemann integral". A Riemann integrable function can have non-removable discontinuity. However a Riemann integrable function can not have an infinite discontinuity (like $1/x$ at $x=0$) because it must be bounded.