Let

G(x)=(x^(2)-12x+35)/(|7-x|)+|x-1|

for

x!=7

\ a) As

x

approaches 7 , the quotient gives an indeterminate form of the type

(\\\\infty )/(\\\\infty )

\\\\infty -\\\\infty

\

1^(\\\\infty )

\

(0)/(0)

\

0\\\\times \\\\infty

\

0^(0)

\ b) Using properties of absolute value, assume

x>7

and simplify the expression

G(x)

.\ FORMATTING: Your answer should not have absolute values or a quotient\ Answer

G(x)=

\ C) Compute the limit as

x

goes to 7 from the right, if it exists.\ FORMATTING: If the limit doesn't exist, write diverges.\

\\\\lim_(x->7^(+))G(x)=

\ (A)\ d) Simplify the expression for

G(x)

when

1

.\ FORMATTING: Your answer should not have absolute values or a quotient\ Answer:

G(x)=

\ \ e) Compute the limit as

x

goes to 7 from the left, if it exists.\ FORMATTING: If the limit doesn't exist, write diverges.\

\\\\lim_(x->7^(-))G(x)=

\ f) Using (c) and (e) above, compute the limit as

x

goes to 7 , if it exists.\ FORMATTING: if the limit doesnt exist, write diverges.\

\\\\lim_(x->7)G(x)=

Let G(x)=7xx212x+35+x1 for x=7 a) As x approaches 7 , the quotient gives an indeterminate form of the type / 1 0/0 0 00 b) Using properties of absolute value, assume x>7 and simplify the expression G(x). FORMATTING: Your answer should not have absolute values or a quotient. Answer G(x)= C) Compute the limit as x goes to 7 from the right, if it exists. FORMATTING: If the limit doesn't exist, write diverges. limx7+G(x)= d) Simplify the expression for G(x) when 1x