Solve limit similar to the well-known one

Question

I am trying to solve the following limit

$\lim_{k \to \infty} \left( 1-\frac{\lambda}{k} \right)^{k}$

I think that the idea is to solve it by using the well-known limit

$\epsilon=\lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^x$

However I cannot transfer one into another

For example, let's take $x=-\frac{k}{\lambda}$ then

$\lim_{-x\lambda \to \infty} \left( 1+\frac{1}{x} \right)^{-x\lambda}$

Here is here I am stuck, because this limit doesn't seem like the well-known one.

I appreciate your help.

Answer 1

You have $$ \left(1+\frac1x\right)^{-x\lambda} =\left[\left(1+\frac1x\right)^x\right]^{-\lambda} $$ As taking exponents is continuous, you can calculate by taking the limit inside the exponent.

Answer 2

You are on the right track: $x=-\frac{k}{\lambda}$ implies $k\to +\infty \Rightarrow x\to -\infty$. So: $$\lim_{-x\lambda \to \infty} \left( 1+\frac{1}{x} \right)^{-x\lambda}=\lim_{x \to -\infty} \left(\left(1+\frac{1}{x} \right)^{x}\right)^{-\lambda}= \left(\lim_{x \to -\infty}\left(1+\frac{1}{x} \right)^{x}\right)^{-\lambda}=e^{-\lambda}.$$ Note: $$\lim_{x\to +\infty} \left(1+\frac 1x\right)^x=\lim_{x\to -\infty} \left(1+\frac 1x\right)^x=e.$$