Find $\lim\limits_{x \to\infty} (\frac {2+3x}{2x+1})^{x+1}$ without L'Hopital

Question

$$ \lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x+1} $$ Not sure how to deal with this, I've tried doing the following $$ \lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x}\cdot \lim_{x\to\infty}\left(\frac{2+3x}{2x+1}\right) $$

Then I tried dividing by $x,$ but my teacher told me it was wrong.

Answer 1

HINT

We have that

$$\left(\frac {2+3x}{2x+1}\right)^{x+1}\sim \left(\frac32\right)^{x+1}$$

Answer 2

It is $\infty,$ because $\frac{2+3x}{2x+1} \to 3/2$ and $x+1 \to \infty.$

Answer 3

The inner fraction converges to $\frac 3 2$, which is larger than 1, so the whole thing goes to $+\infty$.

Answer 4

Write $(\frac {2+3x}{2x+1})^{x+1}=e^{\ln{(\frac {2+3x}{2x+1})^{x+1}}}=$ $ e^{(x+1)\ln(\frac{2+3x}{2x+1}) }$ ...

Answer 5

When $x\to\infty$: $$(\frac{2+3x}{2x+1})^{x+1}=(\frac{3t-1}{2t-1})^t~~~\text{for}~~ (x+1=t)$$ $$=(3/2+\frac{1/2}{2t-1})^t=(3/2+\frac{1/2}{s})^{s}\times3/2$$ where $s=2t-1$ and the last parentheses goes to $+\infty$. This is the same as @gimusi noted.

Answer 6

You get the answer by dividing the numerator and the denominator by x where x is not 0. the numerator witll give you 2/x+3 and the lim when x+1 goes to infinity is 3. The same thing with the denominator will give you 2+1/x and the lim of that is 2 The answer is 3/2.

Answer 7

We have \begin{align} \lim_{x\to\infty}\left(\frac {2+3x}{1+2x}\right)^{1+x} &= \lim_{x\to\infty}\left(\frac {(1+2x)+(1+x)}{1+2x}\right)^{1+x} \\ &= \lim_{x\to\infty}\left(1+\frac {1+x}{1+2x}\right)^{1+x} \\ &= \lim_{x\to\infty}\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{1+x} \\ &= \lim_{x\to\infty}\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x} \\ \end{align} For $ x $ large enough we have $$ 1.3<\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)<3 $$ and $$ 1.3^{1+2x} <\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x} <3^{1+2x} $$ By the sandwich theorem we have to $$ \lim_{x\to \infty}\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}=\infty $$

Answer 8

$$L=\lim_{x\to \infty}\left(\frac{2+3x}{2x+1}\right)^{x+1}$$ we can see that as $x\to \infty$ the $2$ at the top and $1$ at the bottom become arbitary small, such that we can approximate: $$\lim_{x\to \infty}\left(\frac{2+3x}{2x+1}\right)=\lim_{x\to \infty}\left(\frac{3x}{2x}\right)=\frac{3}{2}$$ so it can also be said that: $$\lim_{x\to \infty}\left(\frac{2+3x}{2x+1}\right)^{x+1}=\lim_{x\to \infty}\left(\frac{3}{2}\right)^{x+1}$$ and since $\frac{3}{2}>1$, this tends to infinity, so the limit is not convergent.