The hyperbolic cosine function, denoted

Question

\cosh (x)

, is used to model the shape of a hanging cable (a telephone wire, for example). It is defined as

\cosh (x) = \frac{e^{x} + e^{- x}}{2}

.

a. Determine its end behavior by analyzing $\lim_{x \to \infty} \cosh (x)$ and $\lim_{x \to - \infty} \cosh (x)$ .

Accepted Answer

Welcome back, everyone. The hyperbolic sign function ci of X models the velocity of a particle in hyperbolic motion. It is defined as cinch of X is equal to e the power of X minus e the power of negative x divided by 2. Find the following limits limit of cinch of axis X approaches positive infinity and limit of cinch of X as X approaches negative infinity. So let's focus on the first limit. Limits X approach is positive infinity, and we're going to use the definition of cinch. It's e to the power of X minus e to the power of negative X divided by 2. First of all, we're going to factor out the constant 1/2, and then we're going to rewrite the limit. Limit as X approaches infinity of E to the power of X minus e to the power of negative negative X. Now let's apply the right substitution. We're going to get 1/2 multiplied by E to the power of X simply approaches infinity because we're going to get E to the power of infinity, which approaches infinity, right? So that's infinity, and then we are subtracting 1 divided by e to the power of infinity because e to the power of negative axis 1 divided by e to the power of X, and we are simply dividing 1 by infinity, right, which gives us 0 by definition whenever we are dividing a number by an infinitely large number. Overall, if we subtract 0 from infinity, that's infinity. Infinity divided by 2 is simply infinity. So we have our first limit. And now let's focus on the second one. The main difference is that X. Approaches negative infinity. So we simply want to include negative infinity. And what we're going to do is simply analyze the given limits. So let's make sure that we have negative infinity for every limit. And now in this case, well, we're going to get each the power of negative infinity for the first term. And since we're raising each A negative exponent we can simply rewrite it as e one divided by e to the power of positive infinity for the second term that's eats the power of negative negative infinity. So that means eats the power of positive infinity. And now we can notice that we are simply swapping those that we had in the previous part. One divided by each the power of infinity approaches 0. And then we have eats the power of infinity which approaches infinity. So 0 minus infinity is negative infinity, negative infinity divided by 2 gives us negative infinity. And that said, we have evaluated the two limits, so let's label our final answers. Thank you for watching.

Key Concepts

Hyperbolic Functions

Limits

End Behavior of Functions

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