Evaluate the limit of the indeterminate quotient?

Question

Elijah Abram · Accepted Answer

#lim_(x->0) (sqrt(7-x)-sqrt(7+x))/x = -1/sqrt(7)#

Evaluate the limit:

#lim_(x->0) (sqrt(7-x)-sqrt(7+x))/x#

Rationalize the numerator of the function using the identity: #(a+b)(a-b) = (a^2-b^2)#:

# (sqrt(7-x)-sqrt(7+x))/x = ((sqrt(7-x)-sqrt(7+x))/x)((sqrt(7-x)+sqrt(7+x))/(sqrt(7-x)+sqrt(7+x))) = (7-x-7-x)/(x(sqrt(7-x)+sqrt(7+x))) = -(2x)/(x(sqrt(7-x)+sqrt(7+x)) ) = -2/(sqrt(7-x)+sqrt(7+x))#

Now the limit is determinate:

#lim_(x->0) (sqrt(7-x)-sqrt(7+x))/x = lim_(x->0) -2/(sqrt(7-x)+sqrt(7+x)) = -1/sqrt(7)#

Emilia Aguilar · Answer

#-1/sqrt7#

Here is another way of solving it. Slightly messy.

L'hopital's Rule:

#lim_(x->0)(f(x))/(g(x))=lim_(x->0)(f^'(x))/(g^'(x))# only if the expression is indeterminate .

#f^'(x)# is simply the derivative of #f(x)# with respect to #x#.

Now via L'hopital's Rule (differentiate the numerator and the denominator separately), #lim_(x->0)=(sqrt(7-x)-sqrt(7+x))/x=lim_(x->0)(-1/2(7-x)^(-1/2)-1/2(7+x)^(-1/2))/1=-1/2(1/(sqrt(7-x))+1/(sqrt(7+x)))#

Let #x=0# and you get #-1/sqrt7#

Christopher Agresta · Answer

undefined To evaluate the limit of an indeterminate quotient, we need to apply techniques such as L'Hôpital's rule or algebraic manipulation. Without a specific indeterminate quotient provided, it is not possible to provide a comprehensive answer. Please provide the specific indeterminate quotient for further assistance.