Estimate the limit numerically or state that the limit does not exist: [tex]\[\lim _{x \rightarrow 25} \frac{\sqrt{x}-5}{x-25}\][/tex] Give your answer to at least three decimal places.
Answer (2)
Simplify the expression by multiplying the numerator and denominator by the conjugate: x − 25 x − 5 ⋅ x + 5 x + 5 = x + 5 1 .
Substitute x = 25 into the simplified expression: 25 + 5 1 = 10 1 .
The limit of the function as x approaches 25 is 0.1.
Express the answer to at least three decimal places: 0.100 .
Explanation
Problem Analysis We are asked to estimate the limit of the function x − 25 x − 5 as x approaches 25. Notice that direct substitution leads to an indeterminate form 0 0 , so we need to manipulate the expression to evaluate the limit.
Numerical Estimation Strategy To estimate the limit numerically, we can evaluate the function at values of x close to 25. Let's consider values approaching 25 from both the left and the right.
Simplifying the Expression We can simplify the expression by multiplying the numerator and denominator by the conjugate of the numerator: x − 25 x − 5 = x − 25 x − 5 ⋅ x + 5 x + 5 = ( x − 25 ) ( x + 5 ) x − 25 = x + 5 1 for x = 25 .
Evaluating the Limit Now, we can evaluate the limit by substituting x = 25 into the simplified expression: x → 25 lim x + 5 1 = 25 + 5 1 = 5 + 5 1 = 10 1 = 0.1
Numerical Verification Alternatively, we can use values close to 25 to estimate the limit. From the left, we have values like 24.999, and from the right, we have values like 25.001. Using a calculator, we find that for x = 24.999 , the function value is approximately 0.100001, and for x = 25.001 , the function value is approximately 0.099999. These values are very close to 0.1.
Final Answer Therefore, the limit of the function as x approaches 25 is 0.1. We can express this to at least three decimal places as 0.100.
Examples
In physics, when analyzing the motion of an object, you might encounter a situation where you need to determine the instantaneous velocity at a specific point. This often involves finding the limit of the average velocity as the time interval approaches zero. For example, if the position of an object is given by a function involving a square root, calculating the instantaneous velocity at a particular time might require evaluating a limit similar to the one in this problem. Understanding how to manipulate and evaluate such limits is crucial for accurately determining the object's velocity at that instant.
The limit lim x → 25 x − 25 x − 5 is evaluated by simplifying the expression to x + 5 1 and substituting x = 25 to find it equals 0.1. This can be confirmed numerically by evaluating values close to 25. The final answer, to three decimal places, is 0.100 .
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