How do you evaluate the limit #(sqrt(x+19)-3)/(x+10)# as x approaches #-10#?

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Answer 1 · 2016-10-24T04:35:52

L'Hôpital's rule asserts that #lim_(xto-10)(sqrt(x + 19) -3)/(x + 10) = 1/6#

Because the expression evaluated at the limit is an indeterminate form, specifically #0/0#, then L'Hôpital's rule applies.

Compute the derivative of the denominator

#(d[x + 10])/dx = 1#

Compute the derivative of the numerator:

#(d[sqrt(x + 19) - 3])/dx = 1/(2sqrt(x + 19)#

Assemble a new expression with these:

#lim_(xto-10)(1/(2sqrt(x + 19)))/(1)#

Perform the division by 1:

#lim_(xto-10)1/(2sqrt(x + 19))#

Evaluate at the limit

#lim_(xto-10)1/(2sqrt(x + 19)) = 1/(2sqrt(-10 + 19))= 1/(2sqrt(9)) = 1/6#

The rule asserts that the limit of the original expression is the same:

#lim_(xto-10)(sqrt(x + 19) -3)/(x + 10) = 1/6#