https://file.aiquickdraw.com/mathbot/file/17263097200810qi1zsq4.jpg

Let's analyze the limits shown in your uploaded photo.

1. **For the first limit:**
   \[
   \lim_{x \to 81} \frac{x - 81}{\sqrt{x} - 9}
   \]
   This is an indeterminate form \(\frac{0}{0}\). We can simplify it by multiplying the numerator and the denominator by the conjugate of the denominator:
   \[
   \lim_{x \to 81} \frac{x - 81}{\sqrt{x} - 9} \cdot \frac{\sqrt{x} + 9}{\sqrt{x} + 9} = \lim_{x \to 81} \frac{(x-81)(\sqrt{x} + 9)}{x - 81}
   \]
   Cancel out \(x - 81\) (which is non-zero for all \(x \neq 81\)) to get:
   \[
   \lim_{x \to 81} \sqrt{x} + 9 = \sqrt{81} + 9 = 18
   \]
   Therefore, this limit exists and equals 18.

2. **For the second limit:**
   \[
   \lim_{x \to 6} \frac{x^2 + 9x - 90}{x - 6}
   \]
   Factoring the numerator:
   \[
   x^2 + 9x - 90 = (x - 6)(x + 15)
   \]
   Then the expression becomes:
   \[
   \lim_{x \to 6} \frac{(x-6)(x+15)}{x-6} = \lim_{x \to 6} (x + 15) = 21
   \]
   So, this limit also exists and equals 21.

3. **For the third limit:**
   \[
   \lim_{x \to 3} (x+2) \sin\left(\frac{1}{x-3}\right)
   \]
   The function \(\sin\left(\frac{1}{x-3}\right)\) does not have a limit as \(x\) approaches 3 because \(\frac{1}{x-3}\) approaches infinity or negative infinity, causing the sine function to oscillate between -1 and 1. Thus, this limit does not exist as a real number.

4. **For the fourth limit:**
   \[
   \lim_{x \to 19} \frac{4x - 19}{x - 19}
   \]
   Factoring out 4 from the numerator:
   \[
   \frac{4x - 19}{x - 19} = \frac{4(x - 19) + \frac{19}{4}}{x - 19}
   \]
   This simplifies to:
   \[
   4 + \frac{19}{4(x-19)}
   \]
   As \(x \to 19\), \(\frac{19}{4(x-19)}\) becomes unbounded, so the limit does not exist.

Based on this analysis, the first two limits are correctly marked as having existing real number limits. The third and fourth limits do not exist as real numbers.

Do you want more details on any of these limits or have any questions?

1. Why is the \(\sin\left(\frac{1}{x-3}\right)\) function particularly tricky when finding limits?
2. Can we find a limit for \(\frac{4x-19}{x-19}\) using another approach?
3. How does the multiplication by the conjugate help simplify the first limit?
4. What's the significance of cancelling out terms when calculating limits?
5. Why is the behavior of \(\frac{1}{x-19}\) important in the fourth limit?

**Tip:** Always remember to check if the factor you cancel is non-zero to avoid mathematical errors.

Select all the options where the corresponding limit exists, i.e., the limit is a real number.

Learn how to solve limits involving indeterminate forms and trigonometric oscillations. This problem set explores algebraic simplification techniques and limits involving complex behavior, making it suitable for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation