Math Problem Statement

solve

Find the limits of the sets presented in the image and draw the graph.

Solution

The image you uploaded contains eight limit problems. I will go ahead and solve each of them below:

1. limx2(3x+2)\lim_{x \to 2} (3x + 2)

We can directly substitute x=2x = 2 into the expression: limx2(3x+2)=3(2)+2=6+2=8\lim_{x \to 2} (3x + 2) = 3(2) + 2 = 6 + 2 = 8 Answer: 8\boxed{8}

2. limx2(x21)\lim_{x \to 2} (x^2 - 1)

Substituting x=2x = 2: limx2(x21)=(221)=41=3\lim_{x \to 2} (x^2 - 1) = (2^2 - 1) = 4 - 1 = 3 Answer: 3\boxed{3}

3. limx0(1+1x)\lim_{x \to 0} \left( 1 + \frac{1}{x} \right)

As x0x \to 0, the term 1x\frac{1}{x} grows without bound, leading to: limx0(1+1x)=\lim_{x \to 0} \left( 1 + \frac{1}{x} \right) = \infty Answer: \boxed{\infty} (undefined as xx approaches zero from the right)

4. limx5x1\lim_{x \to 5} \sqrt{x - 1}

Substituting x=5x = 5: limx5x1=51=4=2\lim_{x \to 5} \sqrt{x - 1} = \sqrt{5 - 1} = \sqrt{4} = 2 Answer: 2\boxed{2}

5. limx1x21x1\lim_{x \to 1} \frac{x^2 - 1}{x - 1}

This is an indeterminate form 00\frac{0}{0}, so we factor the numerator: x21x1=(x1)(x+1)x1\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} Canceling out the common factor: limx1(x+1)=1+1=2\lim_{x \to 1} (x + 1) = 1 + 1 = 2 Answer: 2\boxed{2}

6. limx0x23xx\lim_{x \to 0} \frac{x^2 - 3x}{x}

Factoring the numerator: x23xx=x(x3)x\frac{x^2 - 3x}{x} = \frac{x(x - 3)}{x} Canceling xx (for x0x \neq 0): limx0(x3)=03=3\lim_{x \to 0} (x - 3) = 0 - 3 = -3 Answer: 3\boxed{-3}

7. limx3x3x3\lim_{x \to 3} \frac{|x - 3|}{x - 3}

This limit depends on the direction of xx approaching 3:

  • As x3+x \to 3^+ (from the right), x3=x3|x - 3| = x - 3, so: limx3+x3x3=x3x3=1\lim_{x \to 3^+} \frac{|x - 3|}{x - 3} = \frac{x - 3}{x - 3} = 1

  • As x3x \to 3^- (from the left), x3=(x3)|x - 3| = -(x - 3), so: limx3x3x3=(x3)x3=1\lim_{x \to 3^-} \frac{|x - 3|}{x - 3} = \frac{-(x - 3)}{x - 3} = -1

Since the limits from the left and right are not equal, the limit does not exist.

Answer: DNE (Does Not Exist)\boxed{\text{DNE (Does Not Exist)}}

8. limx0xxx\lim_{x \to 0} \frac{|x| - x}{x}

We evaluate the limit by considering the cases:

  • As x0+x \to 0^+ (from the right), x=x|x| = x, so: limx0+xxx=xxx=0\lim_{x \to 0^+} \frac{|x| - x}{x} = \frac{x - x}{x} = 0

  • As x0x \to 0^- (from the left), x=x|x| = -x, so: limx0xxx=xxx=2\lim_{x \to 0^-} \frac{|x| - x}{x} = \frac{-x - x}{x} = -2

Since the limits from the left and right are not equal, the limit does not exist.

Answer: DNE (Does Not Exist)\boxed{\text{DNE (Does Not Exist)}}


Would you like further details on any of these problems? I can also draw graphs for the limits if needed.

Here are five additional questions to extend your understanding:

  1. What happens to a function's behavior when its limit does not exist at a point?
  2. How do piecewise functions affect limit calculations?
  3. What are the conditions for a limit to exist at a certain point?
  4. How can we use factoring to simplify expressions when calculating limits?
  5. What role does one-sided limits play in determining the behavior of functions?

Tip: Always check for indeterminate forms like 00\frac{0}{0} and \frac{\infty}{\infty}, which often require algebraic manipulation or L'Hopital's Rule.

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Math Problem Analysis

Mathematical Concepts

Limits
Algebra
Calculus
Piecewise Functions
Indeterminate Forms

Formulas

Direct substitution for limits
Factoring for indeterminate forms
One-sided limits and absolute value expressions

Theorems

Limit properties
Factorization techniques for limits
One-sided limits theorem
Indeterminate forms: 0/0 and infinity/infinity

Suitable Grade Level

Grade 11-12 or Early College