Guess the value of the limit (if it exists) by evaluating the function at the given numbers.  (It is suggested that you report answers accurate to at least six decimal places.) 
Let
 
f
(
x
)
=
e
3.5
x
-
e
2.9
x
x
.
We want to find the limit
  	 
 
lim
x
→
0
f
(
x
)
=
lim
x
→
0
e
3.5
x
-
e
2.9
x
x
.


Start by calculating the values of the function for the inputs listed in this table.
x
 
f
(
x
)
0.2	Incorrect
0.1	Incorrect
0.05	Incorrect
0.01	
0.001	Incorrect
0.0001	Correct
0.00001	Correct


The given function is:

\[
f(x) = \frac{e^{3.5x} - e^{2.9x}}{x}
\]

We want to find the limit of \( f(x) \) as \( x \to 0 \):

\[
\lim_{x \to 0} \frac{e^{3.5x} - e^{2.9x}}{x}
\]

This limit resembles a derivative form, but for now, let's compute the values of \( f(x) \) for small values of \( x \) to estimate the limit.

### Step-by-step calculation:

1. **For \( x = 0.2 \)**:
\[
f(0.2) = \frac{e^{3.5(0.2)} - e^{2.9(0.2)}}{0.2} = \frac{e^{0.7} - e^{0.58}}{0.2}
\]

2. **For \( x = 0.1 \)**:
\[
f(0.1) = \frac{e^{3.5(0.1)} - e^{2.9(0.1)}}{0.1} = \frac{e^{0.35} - e^{0.29}}{0.1}
\]

3. **For \( x = 0.05 \)**:
\[
f(0.05) = \frac{e^{3.5(0.05)} - e^{2.9(0.05)}}{0.05} = \frac{e^{0.175} - e^{0.145}}{0.05}
\]

4. **For \( x = 0.01 \)**:
\[
f(0.01) = \frac{e^{3.5(0.01)} - e^{2.9(0.01)}}{0.01} = \frac{e^{0.035} - e^{0.029}}{0.01}
\]

5. **For \( x = 0.001 \)**:
\[
f(0.001) = \frac{e^{3.5(0.001)} - e^{2.9(0.001)}}{0.001} = \frac{e^{0.0035} - e^{0.0029}}{0.001}
\]

6. **For \( x = 0.0001 \)**:
\[
f(0.0001) = \frac{e^{3.5(0.0001)} - e^{2.9(0.0001)}}{0.0001} = \frac{e^{0.00035} - e^{0.00029}}{0.0001}
\]

7. **For \( x = 0.00001 \)**:
\[
f(0.00001) = \frac{e^{3.5(0.00001)} - e^{2.9(0.00001)}}{0.00001} = \frac{e^{0.000035} - e^{0.000029}}{0.00001}
\]

### Observing the behavior:
As \( x \) approaches 0, the function converges towards a specific value. This can be verified by directly substituting increasingly smaller values of \( x \). It seems that as \( x \to 0 \), the limit tends to approximately \( 0.6 \), which is the difference in the derivatives of the exponents at \( x = 0 \).

Thus, we estimate that:
\[
\lim_{x \to 0} \frac{e^{3.5x} - e^{2.9x}}{x} \approx 0.6
\]

This conclusion aligns with the correct values given at \( x = 0.0001 \) and \( x = 0.00001 \), indicating that the limit is approximately \( 0.6 \).

---

Would you like further details or have any questions about this?

### Related Questions:
1. What is the derivative of \( e^{ax} \) at \( x = 0 \)?
2. How does this problem relate to the definition of a derivative?
3. Why does the limit approach the difference between the coefficients \( 3.5 \) and \( 2.9 \)?
4. How would we compute this limit using L'Hopital's Rule?
5. What does the function \( f(x) \) represent graphically as \( x \to 0 \)?

### Tip:
When facing limits involving exponential functions, recognize patterns similar to derivatives and use approximations for small values of \( x \).

Guess the value of the limit (if it exists) by evaluating the function at the given numbers. (It is suggested that you report answers accurate to at least six decimal places.) Let f(x) = e^3.5x - e^2.9x / x. We want to find the limit lim x → 0 f(x) = lim x → 0 (e^3.5x - e^2.9x) / x. Start by calculating the values of the function for the inputs listed in this table: 0.2, 0.1, 0.05, 0.01, 0.001, 0.0001, 0.00001.

Discover how to calculate the limit of f(x) = (e^(3.5x) - e^(2.9x)) / x as x approaches 0. This calculus problem involves understanding exponential functions, limits, and derivative concepts. Suitable for college-level students studying calculus.

Math Problem Statement

f ( x )

e 3.5 x

lim x → 0 f ( x )

lim x → 0 e 3.5 x

Solution

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