lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/(1−x)) =?


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Question Number 122721 by bemath last updated on 19/Nov/20

$\lim_{x \to 1} {(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{1 - x}} = ?$
	Answered by Dwaipayan Shikari last updated on 19/Nov/20

	$\lim_{x \to 1} {(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{(1 + \sqrt{x}) (1 - \sqrt{x})}}$ $\lim_{x \to 1} {(\frac{2}{3})}^{\frac{1}{2}} = \sqrt{\frac{2}{3}}$
	Answered by liberty last updated on 19/Nov/20
	$Solve lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/(1−x)) ? Solution : denote h(x)=((1+x)/(2+x)) ; r(x)=((1−(√x))/(1−x)) { ((lim_(x→1) h(x)= (2/3))),((lim_(x→1) r(x)=lim_(x→1) ((1−(√x))/((1+(√x))(1−(√x))))=(1/2))) :} but at finite limits lim_(x→a) h(x) = A>0 , lim_(x→a) r(x)= B the following relation holds true : lim_(x→a) (h(x))^(r(x)) = e^(lim_(x→a) r(x).ln h(x)) = e^(B.ln A) = A^B . Hence lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/(1−x)) = ((2/3))^(1/2) =(√(2/3)) Note: If in handling examples of the form lim_(x→1) (h(x))^(r(x)) it turns out that lim_(x→a) h(x)=1 and lim_(x→a) r(x)=∞ , then the following transformation may be recomended : lim_(x→a) [ h(x)]^(r(x)) = lim_(x→a) {1+[ h(x)−1] }^(r(x)) = lim_(x→a) {[1+(h(x)−1)]^(1/(h(x)−1)) }^(r(x)[h(x)−1 ]) = e^(lim_(x→a) r(x) [h(x)−1 ] ) . ▲$
	$S o l v e \lim_{x \to 1} {(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{1 - x}} ?$ $S o l u t i o n :$ $d e n o t e h (x) = \frac{1 + x}{2 + x}; r (x) = \frac{1 - \sqrt{x}}{1 - x}$ ${\begin{cases} \lim_{x \to 1} h (x) = \frac{2}{3} \\ \lim_{x \to 1} r (x) = \lim_{x \to 1} \frac{1 - \sqrt{x}}{(1 + \sqrt{x}) (1 - \sqrt{x})} = \frac{1}{2} \end{cases}$ $b u t a t f i n i t e l i m i t s \lim_{x \to a} h (x) = A > 0,$ $\lim_{x \to a} r (x) = B t h e f o l l o w i n g r e l a t i o n$ $h o l d s t r u e : \lim_{x \to a} {(h (x))}^{r (x)} = e^{\lim_{x \to a} r (x) . \ln h (x)} = e^{B . \ln A}$ $= A^{B} . H e n c e \lim_{x \to 1} {(\frac{1 + x}{2 + x})}^{\frac{1 - \sqrt{x}}{1 - x}} = {(\frac{2}{3})}^{\frac{1}{2}} = \sqrt{\frac{2}{3}}$ $N o t e : I f i n h a n d l i n g e x a m p l e s o f t h e$ $f o r m \lim_{x \to 1} {(h (x))}^{r (x)} i t t u r n s o u t$ $t h a t \lim_{x \to a} h (x) = 1 a n d \lim_{x \to a} r (x) = \infty,$ $t h e n t h e f o l l o w i n g t r a n s f o r m a t i o n$ $m a y b e r e c o m e n d e d :$ $\lim_{x \to a} {[h (x)]}^{r (x)} = \lim_{x \to a} {1 + [h (x) - 1]}^{r (x)}$ $= \lim_{x \to a} {{[1 + (h (x) - 1)]}^{\frac{1}{h (x) - 1}}}^{r (x) [h (x) - 1]}$ $= e^{\lim_{x \to a} r (x) [h (x) - 1]} . ▴$
	Commented by liberty last updated on 19/Nov/20

	$f o r e x a m p l e f i n d t h e l i m i t :$ $\lim_{x \to \infty} {(\frac{2 x^{2} + 8}{2 x^{2} - 5})}^{6 x^{2} + 5} ?$ $S o l u t i o n : l e t u s d e n o t e :$ $h (x) = \frac{2 x^{2} + 8}{2 x^{2} - 5}; r (x) = 6 x^{2} + 5$ $\lim_{x \to \infty} h (x) = \lim_{x \to \infty} \frac{2 x^{2} + 8}{2 x^{2} - 5} = 1$ $\lim_{x \to \infty} r (x) = \lim_{x \to \infty} (6 x^{2} + 5) = \infty$ $u s e t h e f o r m u l a \lim_{x \to \infty} {(\frac{2 x^{2} + 8}{2 x^{2} - 5})}^{6 x^{2} + 5}$ $= e^{\lim_{x \to \infty} r (x) [h (x) - 1]}; [h (x) - 1 = \frac{13}{2 x^{2} - 5}]$ $= e^{\lim_{x \to \infty} (\frac{13 (6 x^{2} + 5)}{2 x^{2} - 5})} = e^{39} . ✓ ▴$
	Commented by bemath last updated on 19/Nov/20

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