# Finding the limit $\lim_{n\to \infty}\int^n_0 e^{-\lambda x}\mathrm dx$

Find the following limit:

$$\lim_{n\to \infty}\int^n_0 e^{-\lambda x}\mathrm dx$$

for all $\lambda>0$.

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What type of problem is this? Are you posting for your own interest, or as a class problem? – cuabanana Apr 20 '13 at 22:07

let $u = -\lambda x, du = -\lambda dx$ then we have:

$$\lim_{n \to \infty} \int_0^n e^{-\lambda x}dx = \lim_{n\to \infty}\frac{-1}{\lambda}\int_0^{-\lambda n} e^u du$$

Note that the limits of integration change from $x = 0, x=n$ to $u=0, u=-\lambda n$ since we plug in those values to $u = -\lambda x$.

$$= \lim_{n \to \infty} [-\frac{1}{\lambda} e^u ]_0^{-\lambda n} = \lim_{n \to \infty}\frac{1}{\lambda}(1 - e^{-\lambda n}) = \frac{1}{\lambda}(1-\lim_{n \to \infty}e^{-\lambda n}) = \frac{1}{\lambda}$$

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$$\int_0^n e^{-\lambda x}dx=\left[\frac{-1}{\lambda}e^{-\lambda x}\right]_0^n=\frac{1}{\lambda}\left(1-e^{-\lambda n}\right)\underset{\infty}\longrightarrow\frac{1}{\lambda}$$

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Try the substitution $u=-\lambda x$, $du=-\lambda dx$.

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