lim (e^2x -1-2x)/x^2
h→0
A. 2e
B. 2
C. 4
D. Limit doesn't

To find the limit \(\lim_{{h \to 0}} \frac{e^{2h} - 1 - 2h}{h^2}\), we can use the Taylor series expansion of \(e^{2h}\) around \(h = 0\):

\[ e^{2h} \approx 1 + 2h + \frac{(2h)^2}{2!} + \frac{(2h)^3}{3!} + \cdots \]
\[ e^{2h} \approx 1 + 2h + 2h^2 + \frac{8h^3}{6} + \cdots \]
\[ e^{2h} \approx 1 + 2h + 2h^2 + \frac{4h^3}{3} + \cdots \]

Now, substitute this approximation into the given expression:

\[ \frac{e^{2h} - 1 - 2h}{h^2} \approx \frac{(1 + 2h + 2h^2 + \frac{4h^3}{3} + \cdots) - 1 - 2h}{h^2} \]
\[ \approx \frac{2h^2 + \frac{4h^3}{3} + \cdots}{h^2} \]
\[ \approx \frac{2h^2}{h^2} + \frac{\frac{4h^3}{3}}{h^2} + \cdots \]
\[ \approx 2 + \frac{4h}{3} + \cdots \]

As \(h \to 0\), all terms involving \(h\) and higher powers of \(h\) tend to zero, so the limit becomes:

\[ \lim_{{h \to 0}} \left( 2 + \frac{4h}{3} + \cdots \right) = 2 \]

Therefore, the limit is:

\[ \boxed{2} \]

Do you want more details or have any questions?

Here are some related questions for further practice:
1. Evaluate \(\lim_{{h \to 0}} \frac{e^h - 1 - h}{h^2}\).
2. Evaluate \(\lim_{{h \to 0}} \frac{e^{3h} - 1 - 3h}{h^2}\).
3. Use the Taylor series to find \(\lim_{{h \to 0}} \frac{\sin(h) - h}{h^3}\).
4. Evaluate \(\lim_{{h \to 0}} \frac{\cos(h) - 1 + \frac{h^2}{2}}{h^4}\).
5. Evaluate \(\lim_{{h \to 0}} \frac{\ln(1+h) - h}{h^2}\).
6. Use the Taylor series to find \(\lim_{{h \to 0}} \frac{e^{4h} - 1 - 4h}{h^2}\).
7. Find \(\lim_{{h \to 0}} \frac{e^{5h} - 1 - 5h}{h^2}\).
8. Evaluate \(\lim_{{h \to 0}} \frac{e^{6h} - 1 - 6h}{h^3}\).

**Tip:** When dealing with limits involving exponential functions, using the Taylor series expansion can simplify the computation and help in identifying the leading terms that contribute to the limit.

Learn how to evaluate the limit of (e^(2h) - 1 - 2h) / h^2 as h approaches 0 with a detailed step-by-step solution. This problem involves limits and exponential functions, using the Taylor series expansion for accurate computation. Suitable for advanced high school students and those preparing for calculus courses.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation