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Evaluate   lim_(x rarr1^(+))(sqrt(x-1))/(x^(2)-1)

Evaluate limx1+(x1x21)\lim_{x \to 1^{+}}\left(\frac{\sqrt{x-1}}{x^{2}-1}\right)

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Q. Evaluate limx1+(x1x21)\lim_{x \to 1^{+}}\left(\frac{\sqrt{x-1}}{x^{2}-1}\right)
  1. Understand the problem: Understand the problem.\newlineWe need to find the limit of the function x1/(x21)\sqrt{x-1}/(x^2-1) as xx approaches 11 from the right (denoted as x1+x \to 1^+).
  2. Identify the form: Identify the form of the limit.\newlineSubstitute x=1x = 1 into the function to see if the limit can be directly calculated.\newlinelimx1+(x1x21)=11121=00\lim_{x \to 1^+}\left(\frac{\sqrt{x-1}}{x^2-1}\right) = \frac{\sqrt{1-1}}{1^2-1} = \frac{0}{0}\newlineThis is an indeterminate form, so we cannot directly calculate the limit.
  3. Simplify the denominator: Simplify the denominator.\newlineNotice that x21x^2 - 1 is a difference of squares, which can be factored as (x1)(x+1)(x - 1)(x + 1).
  4. Analyze for simplification: Analyze the function for simplification.\newlineWe can't directly simplify the square root in the numerator with the factors in the denominator. However, we can multiply the numerator and denominator by the conjugate of the numerator to remove the square root.
  5. Multiply by the conjugate: Multiply by the conjugate. Multiply the numerator and denominator by the conjugate of the numerator, which is x1\sqrt{x-1}. limx1+x1x21x1x1\lim_{x \to 1^+}\frac{\sqrt{x-1}}{x^2-1} \cdot \frac{\sqrt{x-1}}{\sqrt{x-1}}
  6. Perform the multiplication: Perform the multiplication.\newlineAfter multiplying, we get:\newlinelimx1+x1x21x1\lim_{x \to 1^+}\frac{x-1}{x^2-1}\sqrt{x-1}
  7. Simplify the expression: Simplify the expression.\newlineNow we can cancel out x1x-1 from the numerator and denominator.\newlinelimx1+1(x+1)(x1)\lim_{x \to 1^+}\frac{1}{(x+1)(\sqrt{x-1})}
  8. Evaluate the limit: Evaluate the limit.\newlineNow that the expression is simplified, we can substitute x=1x = 1 to find the limit.\newlinelimx1+1((1+1)11)=1(20)\lim_{x \to 1^+}\frac{1}{((1+1)\sqrt{1-1})} = \frac{1}{(2\cdot 0)}\newlineThis expression seems to suggest a division by zero, which is undefined. However, we must remember that we are taking the limit as xx approaches 11 from the right, not at x=1x = 1. Therefore, x1\sqrt{x-1} is a small positive number, not zero.
  9. Conclude the limit: Conclude the limit.\newlineAs xx approaches 11 from the right, x1\sqrt{x-1} approaches 00, and the denominator approaches 00. The limit of a function as the denominator approaches 00 from the positive side is ++\infty. Therefore, the limit of the function as xx approaches 11 from the right is ++\infty.

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