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Math  /  Calculus

Question(a) lim⁡x→3x−1−29−x2\lim _{x \rightarrow 3} \frac{\sqrt{x-1}-\sqrt{2}}{9-x^{2}}.

Studdy Solution

STEP 1

1. The limit involves a rational expression with square roots in the numerator and a polynomial in the denominator.
2. Direct substitution of x=3 x = 3 results in an indeterminate form 00 \frac{0}{0} .
3. Algebraic manipulation, such as rationalizing the numerator, can help resolve the indeterminate form.

STEP 2

1. Identify the indeterminate form.
2. Rationalize the numerator.
3. Simplify the expression.
4. Evaluate the limit.

STEP 3

Identify the indeterminate form by substituting x=3 x = 3 into the expression:
3−1−29−32=2−29−9=00 \frac{\sqrt{3-1} - \sqrt{2}}{9 - 3^2} = \frac{\sqrt{2} - \sqrt{2}}{9 - 9} = \frac{0}{0}
This is an indeterminate form.

STEP 4

Rationalize the numerator by multiplying the numerator and the denominator by the conjugate of the numerator:
The conjugate of x−1−2 \sqrt{x-1} - \sqrt{2} is x−1+2 \sqrt{x-1} + \sqrt{2} .
Multiply:
x−1−29−x2⋅x−1+2x−1+2 \frac{\sqrt{x-1} - \sqrt{2}}{9-x^2} \cdot \frac{\sqrt{x-1} + \sqrt{2}}{\sqrt{x-1} + \sqrt{2}}

STEP 5

Simplify the expression:
The numerator becomes:
(x−1−2)(x−1+2)=(x−1)−2=x−3 (\sqrt{x-1} - \sqrt{2})(\sqrt{x-1} + \sqrt{2}) = (x-1) - 2 = x - 3
The denominator becomes:
(9−x2)(x−1+2) (9-x^2)(\sqrt{x-1} + \sqrt{2})
The expression is now:
x−3(9−x2)(x−1+2) \frac{x-3}{(9-x^2)(\sqrt{x-1} + \sqrt{2})}

STEP 6

Factor the denominator 9−x2 9 - x^2 as a difference of squares:
9−x2=(3−x)(3+x) 9 - x^2 = (3-x)(3+x)
The expression is now:
x−3(3−x)(3+x)(x−1+2) \frac{x-3}{(3-x)(3+x)(\sqrt{x-1} + \sqrt{2})}

STEP 7

Notice that x−3 x-3 and 3−x 3-x are negatives of each other:
x−3=−(3−x) x-3 = -(3-x)
Cancel x−3 x-3 with 3−x 3-x :
−1(3+x)(x−1+2) \frac{-1}{(3+x)(\sqrt{x-1} + \sqrt{2})}

STEP 8

Evaluate the limit by substituting x=3 x = 3 into the simplified expression:
lim⁡x→3−1(3+x)(x−1+2)=−1(3+3)(3−1+2) \lim_{x \to 3} \frac{-1}{(3+x)(\sqrt{x-1} + \sqrt{2})} = \frac{-1}{(3+3)(\sqrt{3-1} + \sqrt{2})}
=−16(2+2) = \frac{-1}{6(\sqrt{2} + \sqrt{2})}
=−16×22 = \frac{-1}{6 \times 2\sqrt{2}}
=−1122 = \frac{-1}{12\sqrt{2}}
Rationalize the denominator:
=−224 = \frac{-\sqrt{2}}{24}
The limit is −224 \boxed{\frac{-\sqrt{2}}{24}} .

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