Limits and Limits of a Function (Basic Calculus)

Question

Please don't forget to provide a step-by-step and a clear solution, so I can study it well afterwards. Thank you!

1.) Lim (x^2+2x+1)

x-> 1

2.) Construct the table of values for number 1 (note: do not round off the values. write all decimal places of each value as projected in the calculator)

(please imagine these two as a tables, because there's no option here to add one. add a step-by-step with solution and answer the blank on "f(x)")

x l 0 l 0.3 l 0.7 l 0.9 l 0.99 l 0.999 l

f(x) l l l l l l l

x l 2 l 1.5 l 1.3 l 1.1 l 1.01 l 1.001 l

f(x) l l l l l l l

3.) Lim g(x)

x-> -4

4.) Lim g(x)

x-> -2

5.) Lim g(x)

x-> 1

6.) lim

x-> 2 √(x^2+4x+3) / (x^2+2)

7.) What are the limit theorems that you applied to solve number 6?

(multiple choice)

A.) Limit of a Constant

B.) Limit of an Identity

C.) Limit of a Constant Multiple

D.) Limit of a Sum or Difference

E.) Limit of a Product

F.) Limit of a Quotient

G.) Limit of a Power

H.) Limit of a Radical/Root

8.) lim (2x^2-5x+3) / (x-1)

x-> 1

9.) lim (5-2x) / (x-3)

x-> 3+

10.) lim (x^3-4x^2+5x-7) / (4x^3+2x-6)

x-> ∞

11.) Apply down below the Limit Theorem 9.7 in Evaluating the Limit of the Given Function

lim (√9x^2-2) / (2x+7)

x-> +∞

12.) lim 3(4)^5x+3 - 5

x-> 2/5

13.) lim log2 (2x+1)

x-> 3/2

14.) lim (3/4 csc x)

x-> 5π/4

Answer 1

Did you attempt to do any of these problems.

You should!

Then ask specific questions when you get stuck.

If you cannot start any of these problems, you need a tutoring lesson, not the answers to this question.

Remember polynomials are continuous everywhere.

Rational functions are continuous except at zeros of the denominator.

Sin and cos are continuous everywhere, but tangent is discontinuous where cos=0.

In #10, you need to divide numerator and denominator by x³.

Answer 2

I'll go over a few of these.

For number 1, you start limits by plugging in the value that x is approaching into the function. In this case, we plug in 1. We get (1)² + 2(1) + 1 = 4. So the limit is 4

For number 8, you plug in 1 for x and you get 0/0. So here you have to use some algebra

Factor the numerator:

lim (2x-3)(x-1)

^x→1 x -1

The x-1 in the numerator and denominator cancel, leaving us with

lim 2x-3

^x→1

Plugging in 1 again, gives us that the limit equals -1

For number 9, you plug in 3 for x and you get -1/0. A nonzero number divided by zero, can give you either infinity, negative infinity, or DNE (it will do this if it goes to infinity from one side and negative infinity from the other). We're approaching from the positive side, which means that x is greater than 3 by a bit. This means that x - 3 (the denominator) is positive. The numerator gives us -1, so we have a negative divided by a positive, which gives us a negative. So the limit is -∞.