How do you evaluate #sqrt(25-x^2)# as x approaches 5-?

Question

Ezekiel Alexander · Accepted Answer

As #xto5-, sqrt(25-x^2)to0#.

Unless otherwise stated, the graph of #y = sqrt(25-x^2)# means #y=+-sqrt(25-x^2)# and represents the circle with center at the origin and radius = 5. If it is indicated that #y=+sqrt(25-x^2)#, then the graph is the semi-circle above the x-axis, with dead-end-discontinuations, at #(+-5, 0)#, and there exists this limit problem. . .

Emilia Aguilar · Answer

See below.

Here is the reasoning: As #x# approaches #5#, but #x# is less than #5#, #x^2# approaches #25# and is less than #25#. So #25-x^2# is positive and the square root is real. Furthermore, #25-x^2# approaches #0#, So, finally, #sqrt(25-x^2)# approaches #0# More formally, #lim_(xrarr5^-)sqrt(25-x^2) = sqrt(lim_(xrarr5^-)(25-x^2))# # = sqrt(25-lim_(xrarr5^-)(x^2))# # = sqrt(25-(lim_(xrarr5^-)x)^2)# # = sqrt(25-5^2) = 0#

Luke Alvarez · Answer

undefined As x approaches 5-, the expression sqrt(25-x^2) can be evaluated as follows:

1. Substitute the value of x into the expression: sqrt(25-(5-)^2).
2. Simplify the expression inside the square root: sqrt(25-(25-10x+x^2)).
3. Further simplify the expression: sqrt(25-25+10x-x^2).
4. Combine like terms: sqrt(10x-x^2).
5. Since x approaches 5-, substitute this value into the expression: sqrt(10(5-)-(5-)^2).
6. Simplify the expression: sqrt(50-10-25+10x-x^2).
7. Combine like terms: sqrt(25+10x-x^2).

Therefore, as x approaches 5-, sqrt(25-x^2) simplifies to sqrt(25+10x-x^2).