Math

Question Evaluate limx1x1x1\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1} using the provided table of xx and f(x)f(x) values.

Studdy Solution

STEP 1

Assumptions
1. The table provides values of f(x)f(x) for xx approaching 11 from both the left and the right.
2. The function f(x)f(x) is defined as f(x)=x1x1f(x) = \frac{\sqrt{x}-1}{x-1}.
3. The limit we want to evaluate is limx1f(x)\lim _{x \rightarrow 1} f(x).

STEP 2

To evaluate the limit numerically, we should look at the values of f(x)f(x) as xx gets closer and closer to 11 from both sides.

STEP 3

Examine the values of f(x)f(x) as xx approaches 11 from the left (i.e., xx is less than 11).

STEP 4

List the values of f(x)f(x) as xx approaches 11 from the left:
- For x=0.9x = 0.9, f(x)=0.513167019494862f(x) = 0.513167019494862 - For x=0.99x = 0.99, f(x)=0.501256289338003f(x) = 0.501256289338003 - For x=0.999x = 0.999, f(x)=0.500125062539047f(x) = 0.500125062539047 - For x=0.9999x = 0.9999, f(x)=0.500012500624980f(x) = 0.500012500624980

STEP 5

Observe that as xx gets closer to 11 from the left, the values of f(x)f(x) are getting closer to 0.50.5.

STEP 6

Examine the values of f(x)f(x) as xx approaches 11 from the right (i.e., xx is greater than 11).

STEP 7

List the values of f(x)f(x) as xx approaches 11 from the right:
- For x=1.1x = 1.1, f(x)=0.488088481701516f(x) = 0.488088481701516 - For x=1.01x = 1.01, f(x)=0.498756211208895f(x) = 0.498756211208895 - For x=1.001x = 1.001, f(x)=0.499875062461019f(x) = 0.499875062461019 - For x=1.0001x = 1.0001, f(x)=0.499987500624021f(x) = 0.499987500624021

STEP 8

Observe that as xx gets closer to 11 from the right, the values of f(x)f(x) are also getting closer to 0.50.5.

STEP 9

Notice that the values of f(x)f(x) from both sides are converging to the same number, 0.50.5, as xx approaches 11.

STEP 10

Based on the numerical evidence from the table, we can infer that the limit is 0.50.5.
limx1x1x1=0.5\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1} = 0.5
This is a numerical approximation based on the given table of values. For a more rigorous approach, one would typically use algebraic manipulation or L'Hôpital's Rule to find the exact limit.

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