Consider the limit: lim x→0csc(6x)sin(7x).(a) What is the form of the limit?(b) Based on the form, what is the range of possible values the limit can take?(c) You can’t apply L’Hopital’s rule in its current form. Why not? (Hint: L’Hopital’s rule is a theorem...)(d) Manipulate the limit without changing the result so that you can apply L’Hopital’s rule. Explain why, after this manipulation, you can apply L’Hopital’s rule now.(e) Apply L’Hopital’s rule and determine the result of this limit. Verify that the result is within the range you provided in part (b).

evauluate the limit lim x→0 (sin2 (3x))/ x2

Estimate the limit numerically if it exists. (If an answer does not exist, enter DNE.) lim x→−7  x2 + 7/x + 7

Determine limx→3+f(x) given the definition of f(x) below. (If the limit does not exist, enter DNE.)

Estimate the limit numerically or state that the limit doesn't exist lim h→0 cos(9/h)

Calculate the limit algebraically. (If an answer does not exist, enter DNE.)  lim x→−2  x3 + 8 x2 + 7x + 10       If the limit does not exist, say why. (If the limit does exist, so state.) The limit does not exist because the left and right limits exist but are not the same finite value.The limit does not exist because only one of the left or right limits exists as a finite value.     The limit does not exist because one side diverges to ∞ and the other side diverges to −∞.The limit does not exist because it diverges to ∞ or −∞.The limit does exist and is a finite number.

Estimate the limit numerically if it exists. (If an answer does not exist, enter DNE.) lim x→+∞ 6e−6x

If x4 <=ƒ(x)<= x2 for x in [-1, 1] and x2<=ƒ(x) <= x4 for x <-1 and x> 1, at what points c do you automatically know limx-->c ƒ(x)? What can you say about the value of the limit at these points?

Because of their connection with secant​ lines, tangents, and instantaneous​ rates, limits of the form limh→0f(x+h)−f(x)h occur frequently in calculus.   Evaluate this limit for the given value of x and function f.   f(x)=x​, x=19

Find the limit of lim((9+3x)1/2-3)/x as x approaches 0