How can I find the limit of the following? What is the limit of this
$$\lim_{x\to+\infty}\left(1+\frac{4}{2x+3}\right)^x$$
I know that $$\lim_{x\to+\infty}\left(1+\frac{4}{2x}\right)^x$$ will give me $$e^2$$ but the I dont know what to do with the 3.
I have tried bringing them to a common denominator so I got
$$\lim_{x\to+\infty}\left(\frac{2x+7}{2x+3}\right)^x=\lim_{x\to+\infty}\left(e^{x\ln{(\frac{2x+7}{2x+3})}}\right)$$
And then Im stuck again

Let $f(x) = 1 + \frac{4}{ 2x+3}$
And $g(x) = x $
Your limit is of the form $(1)^{\infty}$ , whose value is equal to $$e^{ \lim_{ x \to {\infty}}  {(f(x) - 1 )}{g(x)}} $$
Therefore , $$L = e^{ \lim_{ x \to {\infty}}  {(\frac{4x}{2x+3} )}} $$
Now $$\frac{4x}{2x+3} = {\frac{4x+6}{2x+3} } - \frac{6}{2x+3} = 2 - \frac{6}{2x+3}$$
Therefore , L becomes
$$ L =  e^{ \lim_{x \to \infty} { 2 - \frac{6}{2x+3}} } = e^2$$
