TITLE: Finding the limit of the following sequence: $\left(\frac{2m^2 + m + 4}{2m^2 + 3m + 5}\right)^{3m-3}$
QUESTION [5 upvotes]: So I am trying to find 
  $$ \lim_{m \to \infty} \left(\frac{2m^2 + m + 4}{2m^2 + 3m +5}\right)^{3m-3} $$

I believe this should be represented as $e$ somehow to the power of something, however I can't get to do the algebra in the right way to get to $e$ , I divide by $2m^2$ but then I don't know what to do next.

REPLY [11 votes]: Using the correct hint of the user @gt6989b, I add my solution to your limit.
$$\left(\frac{2m^2 + m + 4}{2m^2 + 3m +5}\right)^{3m-3}=\left[\left(1+\left(\frac{1}{1-\dfrac{2m^2+3m+5}{2m+1}}\right)\right)^{\left(\dfrac{1-\dfrac{2m^2+3m+5}{2m+1}}{1-\dfrac{2m^2+3m+5}{2m+1}}\right)}\right]^{3m-3}$$
$$\left[\left(1+\left(\frac{1}{1-\dfrac{2m^2+3m+5}{2m+1}}\right)\right)^{\left(\dfrac{1}{1-\dfrac{2m^2+3m+5}{2m+1}}\right)}\right]^{\dfrac{6m^2+\ldots}{-2m^2+\cdots}}$$
Hence we have $$\lim_{m \to \infty} \left(\frac{2m^2 + m + 4}{2m^2 + 3m +5}\right)^{3m-3}=e^{-3}=1/e^3$$