Question
Question: $\lim_{x \to 0} \frac{(1+sinx)^{\frac{1}{2024}}+(1-tanx)^{\frac{1}{2024}}}{x^3}$...
limx→0x3(1+sinx)20241+(1−tanx)20241

Does not exist
Solution
To evaluate the limit limx→0x3(1+sinx)20241+(1−tanx)20241, we first evaluate the numerator and the denominator as x approaches 0.
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Evaluate the numerator as x→0:
Let N(x)=(1+sinx)20241+(1−tanx)20241.
As x→0, sinx→sin0=0 and tanx→tan0=0.
So, N(x)→(1+0)20241+(1−0)20241=120241+120241=1+1=2. -
Evaluate the denominator as x→0:
Let D(x)=x3.
As x→0, D(x)→03=0. -
Determine the form of the limit:
The limit is of the form 02. This indicates that the limit will be either +∞, −∞, or does not exist. To determine this, we need to examine the one-sided limits. -
Analyze the one-sided limits:
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Right-hand limit (RHL): As x→0+, x is a small positive number. Therefore, x3 is also a small positive number (x3→0+). The numerator approaches a positive value, 2.
limx→0+x3(1+sinx)20241+(1−tanx)20241=0+2=+∞. -
Left-hand limit (LHL): As x→0−, x is a small negative number. Therefore, x3 is also a small negative number (x3→0−). The numerator approaches a positive value, 2.
limx→0−x3(1+sinx)20241+(1−tanx)20241=0−2=−∞.
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Conclusion:
Since the right-hand limit (+∞) and the left-hand limit (−∞) are not equal, the limit does not exist.
Explanation of the solution:
The numerator approaches 2 as x→0, while the denominator approaches 0. This results in a limit of the form 02. Evaluating the one-sided limits, we find that the right-hand limit is +∞ (since x3→0+) and the left-hand limit is −∞ (since x3→0−). As the one-sided limits are not equal, the overall limit does not exist.