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How do you find the limit of #(sqrtx-2)/(x-4)# as #x->4^-#?

Answer 1

Charles Anctil

#lim_(x->4^-)(sqrtx-2)/(x-4)=1/4#

Since the function, as #x->4^-#, is equal to the indeterminate form: #0/0#

We can apply L'Hopital's Rule:

Therefore:

#lim_(x->4^-)(sqrtx-2)/(x-4)=>0/0#

#=^(L'H) lim_(x->4^-)(1/2x^(-1/2))=1/(2sqrt4)=1/4#

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Answer 2

Axel Alvarez

To find the limit of (sqrt(x)-2)/(x-4) as x approaches 4 from the left (x->4^-), we can substitute the value of x into the expression. By doing so, we get (sqrt(4)-2)/(4-4), which simplifies to (2-2)/(0). Since we have a denominator of 0, this indicates that the limit does not exist.

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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 of #(sqrtx-2)/(x-4)# as #x-&gt;4^-#?

How do you find the limit of #(sqrtx-2)/(x-4)# as #x->4^-#?