as x approaches 2 from the right side, where x>2, f(x) approaches 9/2 (-2/2 + 11/2 = 9/2) even though f(2) = 2(2)-7 = -3. The function is discontinuous.
You probably meant to have some parentheses (sqr(2x^2+4)/(2x+1) or ((sqr2x^2)+4)/(2x+1). if you really meant sqr2x^2 +4/2x + 1 that has no limit as it means (sqr2x^2) + (4/2x) + 1, as x approaches infinity, f(x) then would also approach infinity. So, odds are the problem was really (sqr(2x^2+4)/(2x+1). Then the x terms dominate and f(x) approaches (1/2)sqr2. the x terms cancel except for their coefficients, leaving sqr2/2
limit of f(x)=-4. If you really intended to write the problem with 5+. Possibly you really meant x approaches - 5 . That creates a problem that requires you do more than just plug in the value 5, as that would lead to zero in the denominator. As written just plug 5 in, and simplify (-4(5)-20)/l-5-5l = -40/10 = -4.
the last problem 4) has the same type of step function and limit as 1).
while you don't give the full problem, you probably are asking what is the limit as x approaches 3, from either the left or right side.
From the right side the limit is x/2-4 = 3/2-4 = -5/2. From the left side the limit is -x/2=-3/2 the function is discontinuous