2. Which of the following functions grow faster than e^x as x→...

Question

2. Which of the following functions grow faster than e^x as x→∞? Which grow at the same rate as e^x? Which grow slower?

g. e^(cos(x))

Accepted Answer

Welcome back everyone. Decide which function increases faster as x approaches infinity. U of x equals e to the power of sin x, or v x equals e to the power of x. Aces u of x, bv of x. C, they grow at the same rate, and d cannot be determined. For this problem, let's remember that whenever we are comparing the growth rate of two functions, we basically want to find the limit as x approaches infinity of their ratio. We can take u of x in the numerator and v of x in the denominator. So when we substitute these functions, we are going to write the limit as x approaches infinity of e to the power of sin x divided by e to the power of x. And to identify this limit, we can apply the properties of exponents, that's limit as x approaches infinity of e to the power of sin x minus x. We have the same base different exponents, and we have division. So we're going to subtract the two exponents, right. And we can now analytically understand what this limit is. Sine oscillates between -1 and 1 as x approaches infinity. On average, we can say that we're going to get sin off. X where x approaches infinity, right? So, overall, thinking about the range of sign between -1 and 1, we're just going to have an average value of a constant between -1 and 1. For simplicity, we're gonna just say an average value of 0. However, it doesn't matter what it is, right? We just have to understand that it is between -1 and 1 because sin is oscillating. Because the dominant term is actually negative x as x approaches infinity where subtracting. Infinity from some constant, and constant minus infinity is going to be negative infinity, right? So we're going to get e to the power of negative infinity or e. One, I'm sorry, 1 divided by e to the power of positive infinity using the properties of fractions. And now our denominator is going to approach infinity because that's eat to the power of infinity while the numerator remains a constant. This gives us a limit of 0. A constant divided by an infinitely large number always yields a limit of zero. Whenever we get a limit of 0, it basically means that the denominator function grows faster because it dominates over the numerator function. So v of x grows. Faster, meaning the correct answer for this problem corresponds to the answer of choice B. Thank you for watching

2. Which of the following functions grow faster than e^x as x→∞? Which grow at the same rate as e^x? Which grow slower?g. e^(cos(x))

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2. Which of the following functions grow faster than e^x as x→∞? Which grow at the same rate as e^x? Which grow slower?
g. e^(cos(x))