Compute the limit lim_{x->0} (e^x - e^{-x})/(2x).

Compute the limit lim_{x->0} (e^x - e^{-x})/(2x).

• A. 0
• B. 1/2
• C. 1
• D. 2

Answer: (C)

This limit looks at how the exponential function behaves when we push in opposite directions and then scale by x. It’s natural to rewrite the fraction as sinh x over x, since sinh x = (e^x - e^{-x})/2. For small x, sinh x has the Taylor expansion x + x^3/6 + ..., so sinh x / x = 1 + x^2/6 + ..., which tends to 1 as x approaches 0. A quick alternative is to use L’Hôpital: the top and bottom both go to 0 as x -> 0, so differentiate to get (e^x + e^{-x})/2, and evaluate at x = 0 to obtain (1 + 1)/2 = 1. Thus, the limit is 1.

This limit looks at how the exponential function behaves when we push in opposite directions and then scale by x. It’s natural to rewrite the fraction as sinh x over x, since sinh x = (e^x - e^{-x})/2. For small x, sinh x has the Taylor expansion x + x^3/6 + ..., so sinh x / x = 1 + x^2/6 + ..., which tends to 1 as x approaches 0.

A quick alternative is to use L’Hôpital: the top and bottom both go to 0 as x -> 0, so differentiate to get (e^x + e^{-x})/2, and evaluate at x = 0 to obtain (1 + 1)/2 = 1.

Thus, the limit is 1.