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Exercise

$\lim_{f\to\infty}\left(\frac{e^{2f}-1}{\sin\left(f\right)}\right)$

Step-by-step Solution

1

The limit $\lim_{f\to\infty }\left(\frac{e^{2f}-1}{\sin\left(f\right)}\right)$ is indeterminate, because the limit of trigonometric functions as $f$ approaches infinity is indeterminate

indeterminate

Final answer to the exercise

indeterminate

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  • Integrate by partial fractions
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Main Topic: Limits

In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value.

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