What is the difference between lim_(x→2^− ) and lim


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 66285 by Rio Michael last updated on 12/Aug/19

$W h a t i s t h e d i f f e r e n c e b e t w e e n$ $\underset{x \to 2^{-}}{l i m} a n d$ $\underset{x \to 2^{+}}{l i m}$
	Answered by MJS last updated on 12/Aug/19

	$limit from left respectively right side . the$ $two limits can be different$ $\lim_{x \to 2^{-}} \frac{1}{x - 2} = - \infty$ $\lim_{x \to 2^{+}} \frac{1}{x - 2} = + \infty$ $\lim_{x \to 0^{-}} \sqrt[x]{x^{2}} = + \infty$ $\lim_{x \to 0^{+}} \sqrt[x]{x^{2}} = 0$
	Answered by GordonYeeman last updated on 12/Aug/19

	$t h e l i m i t x \to 2 c o n s i s t s o n t a k i n g$ $a r b i t r a r i l y c l o s e v a l u e s o f 2 (1.99,$ $1.9999, 1.999999 . . .) . x \to 2^{+} i s b a s i c a l l y$ $t h e s a m e b u t t h o s e v a l u e s m u s t b e$ $g r e a t e r t h a n 2 (2.01, 2.000001, e t c) .$ $I n t h e c a s e o f x \to 2^{-} t h e y m u s t b e$ $s m a l l e r$
	Commented by Rio Michael last updated on 12/Aug/19

	$t h a n k y o u s i r s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum