Certainly! Let's evaluate the given limits step by step:
We are given:
[tex]\[ f(x) = \frac{-3 \exp\left(\frac{1}{x}\right) + 4}{-4 \exp\left(\frac{1}{x}\right) + 6} \][/tex]
### Evaluating [tex]\(\lim_{x \rightarrow \infty} f(x)\)[/tex]:
1. Behavior of [tex]\(\exp\left(\frac{1}{x}\right)\)[/tex] as [tex]\(x \rightarrow \infty\)[/tex]:
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\frac{1}{x}\)[/tex] approaches 0. Therefore, [tex]\(\exp\left(\frac{1}{x}\right)\)[/tex] approaches [tex]\(\exp(0)\)[/tex], which is 1.
2. Substitute [tex]\(\exp\left(\frac{1}{x}\right) \approx 1\)[/tex] in [tex]\(f(x)\)[/tex] as [tex]\(x \to \infty\)[/tex]:
[tex]\[
f(x) \approx \frac{-3 \cdot 1 + 4}{-4 \cdot 1 + 6}
\][/tex]
Simplify the expression:
[tex]\[
f(x) \approx \frac{-3 + 4}{-4 + 6} = \frac{1}{2}
\][/tex]
Thus,
[tex]\[
\lim_{x \rightarrow \infty} f(x) = \frac{1}{2}
\][/tex]
### Evaluating [tex]\(\lim_{x \rightarrow -\infty} f(x)\)[/tex]:
3. Behavior of [tex]\(\exp\left(\frac{1}{x}\right)\)[/tex] as [tex]\(x \rightarrow -\infty\)[/tex]:
As [tex]\(x\)[/tex] approaches negative infinity, [tex]\(\frac{1}{x}\)[/tex] again approaches 0. Therefore, [tex]\(\exp\left(\frac{1}{x}\right)\)[/tex] approaches [tex]\(\exp(0)\)[/tex], which is also 1.
4. Substitute [tex]\(\exp\left(\frac{1}{x}\right) \approx 1\)[/tex] in [tex]\(f(x)\)[/tex] as [tex]\(x \to -\infty\)[/tex]:
[tex]\[
f(x) \approx \frac{-3 \cdot 1 + 4}{-4 \cdot 1 + 6}
\][/tex]
Simplify the expression:
[tex]\[
f(x) \approx \frac{-3 + 4}{-4 + 6} = \frac{1}{2}
\][/tex]
Thus,
[tex]\[
\lim_{x \rightarrow -\infty} f(x) = \frac{1}{2}
\][/tex]
### Conclusion
The limits are:
[tex]\[
\boxed{ \left(\frac{1}{2}, \frac{1}{2}\right) }
\][/tex]
