https://r2.erweima.ai/mathbot/image/1725993250403si2vflni.jpg

The given expression is:

\[
\left(\lim_{x \to 16} \frac{x^2 - 100}{x + 10}\right)^5 + \left(\left(\lim_{x \to 0} \frac{e^{6x} - 1}{x}\right)^4 - \lim_{x \to 0} \frac{-\ln(1 + 283(e^{-x} - 1))}{x}\right)
\]

### Step 1: Evaluate \(\lim_{x \to 16} \frac{x^2 - 100}{x + 10}\)

Simplify the expression inside the limit:

\[
\frac{x^2 - 100}{x + 10} = \frac{(x - 10)(x + 10)}{x + 10}
\]

For \(x \neq -10\), we can cancel \(x + 10\):

\[
\frac{(x - 10)(x + 10)}{x + 10} = x - 10
\]

So, the limit is:

\[
\lim_{x \to 16} (x - 10) = 16 - 10 = 6
\]

Thus, 

\[
\left(\lim_{x \to 16} \frac{x^2 - 100}{x + 10}\right)^5 = 6^5 = 7776
\]

### Step 2: Evaluate \(\lim_{x \to 0} \frac{e^{6x} - 1}{x}\)

Using L'Hopital's Rule (since both numerator and denominator approach 0):

\[
\lim_{x \to 0} \frac{e^{6x} - 1}{x} = \lim_{x \to 0} \frac{6e^{6x}}{1} = 6
\]

Thus, 

\[
\left(\lim_{x \to 0} \frac{e^{6x} - 1}{x}\right)^4 = 6^4 = 1296
\]

### Step 3: Evaluate \(\lim_{x \to 0} \frac{-\ln(1 + 283(e^{-x} - 1))}{x}\)

Simplify the expression inside the logarithm:

For small \(x\), \(e^{-x} \approx 1 - x\). Substituting this approximation:

\[
283(e^{-x} - 1) \approx 283(-x) = -283x
\]

Now, use the approximation \(\ln(1 + y) \approx y\) for small \(y\):

\[
-\ln(1 - 283x) \approx -(-283x) = 283x
\]

So, 

\[
\lim_{x \to 0} \frac{-\ln(1 + 283(e^{-x} - 1))}{x} \approx \lim_{x \to 0} \frac{283x}{x} = 283
\]

### Step 4: Combine all parts

Now substitute everything back into the original expression:

\[
7776 + (1296 - 283) = 7776 + 1013 = 8789
\]

### Final Answer

\[
\boxed{8789}
\]

Would you like more details on any step, or do you have any questions?

### Related Questions:
1. What is L'Hopital's Rule and how is it applied?
2. How can approximations like \(\ln(1 + y) \approx y\) be justified?
3. How do we handle limits involving logarithms?
4. What are common techniques for simplifying expressions inside limits?
5. How can Taylor series expansions help in limit evaluation?

**Tip:** Always check if expressions can be simplified before applying complex rules like L'Hopital's; this often saves time and effort!

Evaluate the limit expression: (lim_{x → 16} (x^2 - 100) / (x + 10))^5 + ((lim_{x → 0} (e^{6x} - 1) / x)^4 - lim_{x → 0} (-ln(1 + 283(e^{-x} - 1)) / x)).

This solution explores how to evaluate a complex limit involving both exponential and logarithmic functions using L'Hopital's Rule and limit properties. Learn the step-by-step process, including algebraic simplifications and approximations. Suitable for students in Grades 10-12 or early college level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation