1. ## Limit problem.

Find $\displaystyle \lim_{x \to \infty} $$\displaystyle \left(\frac{x+5}{x+1} \right)^{x+3} 2. Hi. Originally Posted by varunnayudu Find \displaystyle \lim_{x \to \infty}$$\displaystyle \left(\frac{x+5}{x+1} \right)^{x+3}$

It is
$\displaystyle (\frac{x + 5}{x + 1})^{x + 3} = e^{x*LN[(x + 5)/(x + 1)]}*(\frac{x + 5}{x + 1})^3$

and
$\displaystyle lim_{x \to \infty}(\frac{x + 5}{x + 1})^3= 1$

consider lim_{x \to \infty} x*LN[(x + 5)/(x + 1)] = ... = 4

Therefor

$\displaystyle lim_{x \to \infty} e^{x*LN[(x + 5)/(x + 1)]}*(\frac{x + 5}{x + 1})^3 = e^4$

3. Indeed

4. This looks like an 'e' limit.

Remember the useful and famous $\displaystyle \lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$?.

Rewrite:

$\displaystyle \lim_{x\to \infty}\left(1+\frac{4}{x+1}\right)^{x+3}$

Let $\displaystyle t=\frac{4}{x+1}$

$\displaystyle \lim_{t\to 0}\left(1+t\right)^{\frac{4}{t}+2}$

$\displaystyle \lim_{t\to 0}\left(1+t\right)^{\frac{4}{t}}\cdot\lim_{t\to 0}\left(1+t\right)^{2}$

Now, it is obvious the right limit goes to 1.

So, we have $\displaystyle \left[\lim_{t\to 0}\left(1+t\right)^{\frac{1}{t}}\right]^{4}$

See what the inside limit is?. e.

Therefore, we get $\displaystyle e^{4}$

5. Hello, varunnayudu!

A variation of Rapha's solution . . .

Find: .$\displaystyle \lim_{x \to \infty}\left(\frac{x+5}{x+1} \right)^{x+3}$

We have: .$\displaystyle y \;=\;\left(\frac{x+5}{x+1}\right)^{x+1}\left(\frac {x+5}{x+1}\right)^2$

Use long division on the first fraction.
In the second fraction, divide top and bottom by $\displaystyle x.$

. . $\displaystyle y \;=\;\left(1 + \frac{4}{x+1}\right)^{x+1}\left(\frac{1 + \frac{5}{x}}{1 + \frac{1}{x}}\right)^2$

$\displaystyle \text{Then: }\;\lim_{x\to\infty}y \;=\;\underbrace{\lim_{x+1\to\infty}\left(1 + \frac{4}{x+1}\right)^{x+1}}_{\text{This is }e^4}\cdot\underbrace{ \lim_{x\to\infty}\left(\frac{1 + \frac{5}{x}}{1 + \frac{1}{x}}\right)^2}_{\text{This is 1}} \;=\;e^4$