lim x -> infinity (e^-x + 9cos2x)
To find the limit of the expression, we can consider the behavior of each term separately as x approaches infinity.
1. The term e^(-x) approaches 0 as x tends to infinity since the exponential function decays rapidly.
2. The term 9cos(2x) oscillates between -9 and 9 without a clear trend as x increases. It does not approach a specific value.
Therefore, as x approaches infinity, the overall expression e^(-x) + 9cos(2x) tends to 0 + 0 = 0.
Hence, the limit of the expression as x approaches infinity is 0.
To find the limit of the given function as x approaches infinity, we can analyze the behavior of each term separately.
Let's consider the term e^-x first. As x approaches infinity, the exponential term e^-x will tend to zero. This is because the exponential decay function approaches zero as x goes to infinity. Therefore, the contribution of e^-x to the overall limit will be zero.
Now, let's focus on the second term, 9cos2x. The cosine function oscillates between -1 and 1. As x approaches infinity, cos2x will continue to oscillate indefinitely. However, the amplitude of the oscillation is 9, so the term 9cos2x will be bounded between -9 and 9.
Combining the contributions from both terms, we have 0 (from e^-x) plus a bounded term (9cos2x). The sum of these two terms is still bounded between -9 and 9.
Therefore, the limit of the given function as x approaches infinity is a bounded value between -9 and 9.
In summary:
lim x -> infinity (e^-x + 9cos2x) = (lim x -> infinity e^-x) + (lim x -> infinity 9cos2x) = 0 + bounded term = bounded value between -9 and 9.
To find the limit of the function as x approaches infinity, we will evaluate the limits of each term separately.
1. First, consider the term e^(-x). As x approaches infinity, the exponential function e^(-x) approaches 0 since the exponential term is decreasing exponentially.
lim x -> infinity (e^(-x)) = 0
2. Next, consider the term 9cos(2x). The cosine function oscillates between -1 and 1 as x increases indefinitely. Since x approaches infinity, cos(2x) does not have a limit. However, the product with the constant 9 allows us to conclude that the term oscillates between -9 and 9.
lim x -> infinity (9cos(2x)) = oscillates between -9 and 9
3. Now, let's combine both terms:
lim x -> infinity (e^(-x) + 9cos(2x))
Since the first term approaches 0 and the second term oscillates between -9 and 9, the overall limit does not exist.