Computing $\lim_{n \to \infty} P \left( \left[ \frac{1}{4}, 1 - e^{-n} \right] \right) $ given the uniform distribution on [0,1]

Question

I am currently reading a textbook where I was introduced to the continuity theorem. One of the questions at the end of the chapter is asking me to compute (with proof) $$\lim_{n \to \infty} P \left( \left[ \frac{1}{4}, 1 - e^{-n} \right] \right) $$ given the uniform distribution on [0,1].

I am not exactly sure how to even start, and would appreciate any kind of help and/or guidance (e.g., resources).

Answer 1

The probability of an interval under the uniform distribution is simply the length of the interval:

$$\lim_{n \to \infty} P \left( \left[ \frac{1}{4}, 1 - e^{-n} \right] \right) = \lim_{n \to \infty} 1 - e^{-n} - \frac{1}{4} = \lim_{n \to \infty} \frac{3}{4} - e^{-n} =\frac{3}{4} $$