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How do you find the limit #lim_(x->2^+)sqrt(2-x)# ?

Answer 1

#x\to 2^+# means that #x# is approaching 2 from the right, that is, #x>2#. But if #x>2# then #2-x<0# so that #2-x# is not in the domain of the square root function...So #\lim_{x\to 2^+}\sqrt(2-x)# does not exist , while from the left: #\lim_{x\to 2^-}\sqrt(2-x)=sqrt(0)=0#

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Answer 2

Emily Ammerman

To find the limit as x approaches 2 from the right (x->2+), of the function sqrt(2-x), we substitute the value 2 into the function. This gives us sqrt(2-2), which simplifies to sqrt(0). The square root of 0 is 0. Therefore, the limit as x approaches 2 from the right of sqrt(2-x) is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit #lim_(x-&gt;2^+)sqrt(2-x)# ?

How do you find the limit #lim_(x->2^+)sqrt(2-x)# ?