Calculus

3. Find the value of the following limits if they exist. Graph with calculator to help with a - d. (These four are 4 pts each.) a) lim_{x->0^+} sqrt(x) = b) lim_{x->-4} sqrt(16 - x^2) = c) lim_{x->4^-} 1/(x - 4) = d) lim_{x->3^-} |x - 3|/(x - 3) = SHOW ALL WORK TO JUSTIFY your answers to e - g. (These three are 5 pts each) e) lim_{x->1} (x^2 - x)/(2x^2 + 5x - 7) (use factoring)

Step-by-step solution with explanation

Final Answer

a) | b) | c) (DNE) | d) | e)

Step-by-step solution

Evaluate limit (a): right-hand limit

We approach x = 0 from the right (positive side). Since sqrt(x) is continuous for x ≥ 0, we simply plug in x = 0.

Compute limit (a) result

As x gets closer to 0 from the right, sqrt(x) gets closer to 0. The limit exists and equals 0.

Evaluate limit (b): plug in x = -4

The function sqrt(16 - x^2) is defined when 16 - x^2 ≥ 0, which means -4 ≤ x ≤ 4. Since x = -4 is the endpoint of the domain, we check the value there.

Compute limit (b) result

Plugging in x = -4 gives 0 under the square root. The limit exists and equals 0 (approaching from inside the domain, i.e., from the right).

Evaluate limit (c): left-hand limit near x = 4

As x approaches 4 from the left, (x - 4) is a small negative number. Dividing 1 by a tiny negative number gives a very large negative value.

Compute limit (c) result

Since the denominator approaches 0 from the negative side, the fraction goes to negative infinity. The limit does not exist (DNE) as a finite number — it equals .

Evaluate limit (d): left-hand limit of |x-3|/(x-3)

When x approaches 3 from the left, (x - 3) is negative. The absolute value |x - 3| = -(x - 3) for negative values. So we simplify.

Compute limit (d) result

For x just less than 3, the expression always equals -1. So the left-hand limit is -1.

Evaluate limit (e): factor numerator and denominator

Direct substitution gives 0/0, which is indeterminate. We factor both the top and bottom to cancel common factors.

Factor numerator and denominator

and

For the denominator, we need two factors of (2)(-7) = -14 that add to 5: those are 7 and -2. So 2x² + 5x - 7 = (2x+7)(x-1).

Cancel the common (x - 1) factor

Since x ≠ 1 in a limit (we never actually reach x = 1), we can legally cancel (x-1) from top and bottom.

Substitute x = 1 into simplified expression

Now direct substitution works. Plug in x = 1 to get the final answer for limit (e).

Understanding this problem

Learning Insight

Limits describe what a function approaches, not what it equals at that point. For 0/0 indeterminate forms, factoring lets you remove the problematic factor so substitution works. One-sided limits (like 0⁺ or 4⁻) tell you to approach from only one direction, which matters when the function behaves differently on each side.

Quick Tip

For |x - a|/(x - a): this expression always equals +1 when x > a and -1 when x < a. Memorize this pattern — it comes up on almost every limit test involving absolute values.

Common Mistake

On the absolute value limit (d), students forget that |x - 3| = -(x - 3) when x < 3, and they write |x-3|/(x-3) = 1 instead of -1. Always check the sign of (x - a) based on which side you are approaching from.