Let L 1 = lim x → 0 f ( x ) x and L 2 = lim x → 0 f ( x ) 4 − 4 e x .
Rewrite L 2 as lim x → 0 f ( x ) 4 ( 1 − e x ) .
Multiply and divide the numerator of L 2 by x to get L 2 = lim x → 0 x 4 ( 1 − e x ) ⋅ f ( x ) x .
Since lim x → 0 x 1 − e x = − 1 , we have L 2 = − 4 L 1 , so lim x → 0 f ( x ) 4 − 4 e x = − 4 L 1 .
Explanation
Problem Setup We are given two limits:
x → 0 lim f ( x ) x
x → 0 lim f ( x ) 4 − 4 e x
We need to find the values of these limits.
Define Limits Let's denote the first limit as L 1 and the second limit as L 2 :
L 1 = x → 0 lim f ( x ) x
L 2 = x → 0 lim f ( x ) 4 − 4 e x
We can rewrite L 2 as:
L 2 = x → 0 lim f ( x ) 4 ( 1 − e x )
Rewrite the Second Limit Now, let's multiply and divide the numerator of L 2 by x :
L 2 = x → 0 lim x 4 ( 1 − e x ) ⋅ f ( x ) x
We know that lim x → 0 x 1 − e x = − 1 . Therefore,
L 2 = x → 0 lim 4 ( − 1 ) ⋅ f ( x ) x = − 4 x → 0 lim f ( x ) x
So, we have:
L 2 = − 4 L 1
Final Relationship This shows that the second limit is -4 times the first limit. Without more information about the function f ( x ) , we cannot determine the exact values of L 1 and L 2 . However, we have found the relationship between them.
Therefore, if lim x → 0 f ( x ) x = L 1 , then lim x → 0 f ( x ) 4 − 4 e x = − 4 L 1 .
Final Answer The relationship between the two limits is:
x → 0 lim f ( x ) 4 − 4 e x = − 4 x → 0 lim f ( x ) x
So, if lim x → 0 f ( x ) x = L 1 , then lim x → 0 f ( x ) 4 − 4 e x = − 4 L 1 .
Examples
In physics, when analyzing the behavior of systems near equilibrium, you might encounter limits involving unknown functions. For instance, if f ( x ) represents a damping force dependent on displacement x , understanding the relationship between lim x → 0 f ( x ) x and lim x → 0 f ( x ) 4 − 4 e x can help determine the system's stability and response to small disturbances. This type of analysis is crucial in designing stable control systems and predicting the behavior of physical systems.