limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

Find the following limit:
lim_(x->∞) ((x + 2)/(x - 2))^(x + 1)

lim_(x->∞) ((x + 2)/(x - 2))^(x + 1) = lim_(x->∞) exp(log(((x + 2)/(x - 2))^(x + 1))) = lim_(x->∞) exp((x + 1) log((x + 2)/(x - 2))):
lim_(x->∞) exp(log((x + 2)/(x - 2)) (x + 1))

lim_(x->∞) exp((x + 1) log((x + 2)/(x - 2))) = exp(lim_(x->∞) (x + 1) log((x + 2)/(x - 2))):
exp(lim_(x->∞) log((x + 2)/(x - 2)) (x + 1))

lim_(x->∞) (x + 1) log((x + 2)/(x - 2)) = lim_(x->∞) x log((x + 2)/(x - 2)) (x + 1)/x:
exp(lim_(x->∞) x log((x + 2)/(x - 2)) (x + 1)/x)

By the product rule, lim_(x->∞) x (x + 1)/x log((x + 2)/(x - 2)) = (lim_(x->∞) (x + 1)/x) (lim_(x->∞) x log((x + 2)/(x - 2))):
exp(lim_(x->∞) x log((x + 2)/(x - 2)) lim_(x->∞) (x + 1)/x)

The leading term in the denominator of (x + 1)/x is x. Divide the numerator and denominator by this:
exp(lim_(x->∞) x log((x + 2)/(x - 2)) lim_(x->∞) (1 + 1/x)/1)

The expression 1/x tends to zero as x approaches ∞:
exp(lim_(x->∞) x log((x + 2)/(x - 2)))

To prepare the product x log((x + 2)/(x - 2)) for solution by l'Hôpital's rule, write it as (log((x + 2)/(x - 2)))/(1/x):
exp(lim_(x->∞) (log((x + 2)/(x - 2)))/(1/x))

Applying l'Hôpital's rule, we get that
lim_(x->∞) (log((x + 2)/(x - 2)))/(1/x) | = | lim_(x->∞) ( d/( dx) log((x + 2)/(x - 2)))/( d/( dx)(1/x))
 | = | lim_(x->∞) (((x - 2) (1/(x - 2) - (x + 2)/(x - 2)^2))/(x + 2))/(-1/x^2)
 | = | lim_(x->∞) (4 x^2)/((x - 2) (x + 2))
exp(lim_(x->∞) (4 x^2)/((x - 2) (x + 2)))

Replace -2 by 0 in 1/(x - 2):
exp(lim_(x->∞) (4 x^2)/(x (x + 2)))

The leading term in the denominator of (4 x)/(x + 2) is x. Divide the numerator and denominator by this:
exp(lim_(x->∞) 4/(1 + 2/x))

The expression 2/x tends to zero as x approaches ∞:
Answer: |e^4

Hi Kulki   

It is nice to meet you :)

 

limit( ((((x+2)/(x-2)))^(x+1)), x=infinity )

 

Mmm  it is good that you use lots of brackets but maybe if some of them were square brackets it might be easier to make sense of.   Not to worry, lots of backets is good.

 

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}\)

 

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{ln\left[ \frac{x+2}{x-2}       \right]^{x+1}}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{(x+1)ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{x(1+\frac{1}{x})ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{x(\frac{x+1}{x})ln\left[ \frac{x+2}{x-2}       \right]}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{xln\left[ \frac{x+2}{x-2}      \right]*(\frac{x+1}{x}) }\\ \text{The limit as x tends to infinity of (x+1)/x = 1 so}\\ =\displaystyle\lim_{x\rightarrow\infty}\;e^{xln\left[ \frac{x+2}{x-2}      \right]}\\ =e^\left[{\displaystyle\lim_{x\rightarrow\infty}\;{xln\left[ \frac{x+2}{x-2}       \right]}}\right]\\ \)

 

Now  I will look at the nw limit

\(\displaystyle\lim_{x\rightarrow\infty} \frac{ln\left[ \frac{x+2}{x-2} \right]}{\frac{1}{x}}\\ \qquad \text{The numerator and the denominator both tend to 0 so I can use L'Hopital's rule}\\ \qquad \frac{d}{dx}\;ln\left[ \frac{x+2}{x-2}\right]=\frac{(x-2)-(x+2)}{(x-2)^2}=\frac{-4}{(x-2)^2}\\ \qquad \frac{d}{dx}\;x^{-1}=-x^{-2}=\frac{-1}{x^2}\\ \qquad =\displaystyle\lim_{x\rightarrow\infty}\left[ \frac{-4}{(x-2)^2}\div \frac{-1}{x^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\left[ \frac{4x^2}{(x-2)^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{x^2}{(x-2)^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{x^2}{x^2-4x+4} \;\;\;\frac{\div x^2}{\div x^2}\right]\\ \qquad=\displaystyle\lim_{x\rightarrow\infty}\;4\left[ \frac{1}{1-\frac{4}{x}+\frac{4}{x^2}} \right]\\ \qquad=4\left(\frac{1}{1-0+0}\right)\\ \qquad=4\\ \text{so the original limit}=e^4 \)

 

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[ \frac{x+2}{x-2}       \right]^{x+1}=e^4\)