Limit Has No Limits!

We can write the expression under limit as, $\frac{((1+\frac{3}{x})(1+\frac{5}{x})(1+\frac{7}{x})(1+\frac{9}{x}))^{\frac{1}{4}}-1}{\frac{1}{x}}$ Now putting $\frac{1}{x}=y$ so that as $x\to\infty, y\to 0$ We get, $L=\lim_{y\to 0}\frac{((1+3y)(1+5y)(1+7y)(1+9y))^{\frac{1}{4}}-1}{y}$ The limit has now transformed to $\frac{0}{0}$ form, thereby giving way to a smooth application of L' Hospital to get to the final result.

After the first differentiation of the numerator & denominator, we can put the limits. There is no need to differentiate further as the expression will not result in any of the indeterminate forms.

I'm jumping the differentiation stage which is easy.

We've the final result,

$L=\frac{\frac{1}{4}(1)^{\frac{1}{4}}•(9•1+7•1+5•1+3•1)}{1}=\boxed6$

We can also solve this limit without L'Hopital's rule. First let

$A = (x + 3)(x + 5)(x + 7)(x + 9) = x^{4} + 24x^{3} + 206x^{2} + 744x + 945$ .

Multiply the expression $A^{\frac{1}{4}} - x$ by $1$ in the form $\dfrac{A^{\frac{1}{4}} + x}{A^{\frac{1}{4}} + x}$ to end up with $\dfrac{A^{\frac{1}{2}} - x^{2}}{A^{\frac{1}{4}} + x}$ .

Now multiply this last expression by $1$ in the form $\dfrac{A^{\frac{1}{2}} + x^{2}}{A^{\frac{1}{2}} + x^{2}}$ to end up with

$\dfrac{A - x^{4}}{(A^{\frac{1}{4}} + x)(A^{\frac{1}{2}} + x^{2})} = \dfrac{24x^{3} + 206x^{2} + 744x + 945}{A^{\frac{3}{4}} + A^{\frac{1}{4}}x^{2} + A^{\frac{1}{2}}x + x^{3}} = \dfrac{x^{3}(24 + \frac{206}{x} + \frac{744}{x^{2}} + \frac{945}{x^{3}})}{x^{3}B^{\frac{3}{4}} + x^{3}B^{\frac{1}{4}} + x^{3}B^{\frac{1}{2}} + x^{3}}$ ,

where $B = 1 + \frac{24}{x} + \frac{206}{x^{2}} + \frac{744}{x^{3}} + \frac{945}{x^{4}}$ .

After canceling the $x^{3}$ terms from top and bottom, and after noting that each of $B^{\frac{3}{4}}, B^{\frac{1}{2}}$ and $B^{\frac{1}{4}}$ go to $1$ as $x \rightarrow \infty$ , we find that the original limit goes to

$\dfrac{24}{4} = \boxed{6}$ as $x \rightarrow \infty$ .

The first part of our limit is written as the geometric mean of four terms which are (x+3),(x+5),(x+7) and (x+9). As x tends to infinity we get, (x+3)=(x+5)=(x+7)=(x+9). That's why we can rewrite the given geometric mean as the arithematic mean of those four terms. from here we get that part of our limit as (x+6). putting this value in the original question(x+6-x), we get answer 6

Let

F(X)=x^1/4 Now it takes as an argument the thing (X+3)(X+5)(X+7)(X+9)=T Now as X goes to infinity The function F(T)=something Let's approximate it T = X^4+ 24X^3+ lower powers Now derivative of F Is 1/4x^(3/4)=z Note let's see T go up to x^4 So F reaches X Now here the derivative is z(x^4) And for this rate of change next we multiply the increment Of x with the derivative (1/4 1/x^3) 24x^3=6 So the increment after the function reaches X is 6

So answer is( X+6)-X=6

We use the first order taylor expansion of $(1+y)^\alpha$ as $y \rightarrow 0$ : $(1+y)^\alpha = 1+ \alpha y + y^2M_\alpha(y),$ where $M_\alpha(x)$ is a function bounded around $0$ . Let $f(x)$ be the expression which limit we are looking for. Factoring the whole expression by $x$ and applying the formula above gives us : $f(x) = x\left( \prod_{k=1}^4 \left(1+\frac{2k+1}{x} \right)^{1/4} - 1 \right)$ $= x \left(\prod_{k=1}^4 \left(1+\frac{(2k+1)}{4x}+\frac{(2k+1)^2M_{1/4}(\frac{2k+1}{x})}{x^2}\right) - 1\right)$ $=x \left( 1 + \frac{1}{4x}\sum_{k=1}^3 (2k+1) + \frac{1}{x^2} M(x) -1 \right) = \frac{24}{4} + \frac{1}{x}M(x) ,$ $M(x)$ being a function bounded for $x\rightarrow \infty$ . We therefore have the result : $\frac{24}{4} = \fbox{6}$