Evaluate the limit: limx→∞(tan−1(3x))2(e5x4/3−1)tan(5x1/3)loge(1+3x2) is equal to:
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Step-by-Step Solution
Step 1: Simplify terms using standard limits
When evaluating limits as x→∞, we often use standard limit approximations. For very large x, we can simplify terms like tan−1(3x) and e5x4/3−1.
Step 2: Approximate terms for large x
As x→∞, 3x→∞, so tan−1(3x) approaches 2π. Also, e5x4/3 becomes much larger than 1, so e5x4/3−1≈e5x4/3. For loge(1+3x2), as x→∞, 3x2 dominates 1, so loge(1+3x2)≈loge(3x2).
Step 3: Substitute approximations into the limit expression
Now we substitute these approximations back into the original limit expression. The denominator simplifies to (2π)2e5x4/3.
Step 4: Analyze the behavior of the terms
As x→∞, 5x1/3→∞. The tangent function tan(y) oscillates between −∞ and ∞ as y→∞. It does not approach a single finite value.
Step 5: Determine the limit
The numerator contains an oscillating term tan(5x1/3) and a term (loge(3)+2loge(x)) that goes to ∞. The denominator e5x4/3 goes to ∞ much faster than the numerator. However, due to the oscillating nature of the tangent function, the limit does not exist.