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

2 Answers
A. S. Adikesavan
Apr 15, 2016

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

Explanation:

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. . .

Jim H
Apr 15, 2016

See below.

Explanation:

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#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
4371 views around the world
You can reuse this answer
Creative Commons License