How can I find the limit of the following?

Question

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

What have you tried? Do you know about the natural logarithm function and its properties? — Sammy Black, Sep 13 '21 at 22:12
Hint: $$ \left(1+\frac{4}{2x+3}\right)^x = \left(1+\frac{2}{x+\frac32}\right)^x = \left(1+\frac{2}{x+\frac32}\right)^{x+\frac32} \left(1+\frac{2}{x+\frac32}\right)^{-\frac32} $$ — md2perpe, Sep 13 '21 at 22:30
Or also $$\left(1+\frac{4}{2x+3}\right)^x=\left[\left(1+\frac{4}{2x+3}\right)^{\frac{2x+3}{4}}\right]^{\frac{4x}{2x+3}}$$ — user, Sep 13 '21 at 22:38
@SammyBlack I think that now you can add your answer again since the asker gave some context about its effort. — user, Sep 13 '21 at 22:45

score 3 · Accepted Answer · answered Sep 13 '21 at 22:50

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

$$ L = e^{ \lim_{x \to \infty} { 2 - \frac{6}{2x+3}} } = e^2$$

How can I find the limit of the following?

1 Answers1