Calculus limit, anyone?

Question

I want to confirm if it's true that each time we get a limit equals to infinity, it means that our limit doesnt exist?
Also, assuming x approaches 0
I'm given the
lim [(3 cos 3x+a+2bx)/2x] = 4 ........(1)
It is known that the limit of 2x when x approaches 0 is 0.
And thus the
lim (3 cos 3x+a+2bx) = 0 ..............(2)
Can anyone explains to me how to get from (1) to (2) please?
p/s: Question asked to find values of a and b.
Thank you!

Answer 1

a = -3 & b = 4
________________________________________________
Explanation:

Using limit equation (1):

Since you already know that the denominator approaches 0, but you are given a valid limit (not ∞ or -∞), then the limit would be indeterminate at this point.
The numerator also approaches 0 (known by looking at limit equation (2)), thus fitting the equation to the indeterminate form 0/0.

Now, because you have the indeterminate form 0/0, you can apply L'Hopital's rule, making the new limit equation (-9sin(3x)+2b)/2.
Given that the limit approaches 4 when x approaches 0, pluggin in 0 into that new equation should give you 4

Therefore, (-9sin(3*0)+2b)/2 = 4
at this point, you can solve for b
(-9sin(0)+2b)/2 = 4
(-9(0)+2b)/2 = 4
2b/2 = 4
b = 4
_____________________________________________

Now to solve for a using limit equation (2):

You're given that 3cos(3x)+a+2bx approaches 0 when x approaches 0. Plug in the already known value of b=4.

Therefore, 3cos(3x)+a+8x should approach 0 when x approaches 0. Now, simply plug in 0 for x and you should be able to find the value of a.

3cos(3*0)+a+8(0) = 0
3cos(0)+a+0 = 0
3(1)+a = 0
a = -3

Therefore, a = -3 & b = 4
______________________________________________________

P.S: For your first question, the answer is yes

Even if a limit approaches ∞ or -∞, it technically "does not exist" because it does not actually approach a value. Rather, it approaches ∞ or -∞, merely concepts, not actual values.

Answer 2

Saying that limit is infinity is another way to say that the function is not bounded
As x=>0 3cos3x +a+ 2bx => 3+a and if a is not -3 the given limit to you is infinity.

If a=-3 you have the limit of [3*(cos 3x-1)/2x +b]

As 3cos3x-1 is equivalent for x=>0 to -9/2x^2 the term

3(cos3x-1)/2x =>0.
So, in this case the limit is b So if the limit has to be 4 a=-3 and b=4

Answer 3

If you have the fraction

lim as x→0 of [(3 cos 3x+a+2bx)/2x] = 4
If the denominator = 0 and the numerator does not, then the limit is either ∞ or -∞. Therefore since the limit is 4

lim as x→0 of (3 cos 3x+a+2bx) = 0.

Answer 4

hi, I. f ' (x) < 0 for all x >= 0 is real employing reality the graph is reducing on that section so the slope is adverse. II. shrink (as x --> effective infinity) of f ' (x) = 0 is real once you talk it incredibly is getting closer and closer to being horizontal with a slope of 0. III. shrink (as x--> adverse infinity) of f '(x) = 3 is pretend employing reality employing reality the graph is going left, it incredibly is likewise getting closer and closer to being horizontal with a slope of 0 instead of three. the incredible answer is I and II in basic terms. <==answer i desire that helps!! :-)