Find the limit. lim x → 5 x − 5 x − 5 \lim_{x\rarr5}\frac{x-5}{\sqrt{x}-\sqrt5} lim x → 5 x <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> − 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> x − 5

Here we're asked to find the limit of x minus 5 over the square root of x minus the square root of 5 as x approaches 5. Now if I were to go ahead and plug 5 in here that would make my denominator 0 and already since I see these radicals in this rational function I know that I need to multiply by the conjugate. Now here my my radicals are actually in my denominator so that means I need to take the conjugate here. You're taking the conjugate wherever your radicals are. Now here I have the square root of x minus the square root of 5. So taking the conjugate and multiplying by it that's going to be the square root of x plus the square root of 5, just reversing that sign in between my two terms. So if I'm multiplying by that in the denominator, I need to multiply by that in the numerator as well, the square root of x plus the square root of 5. Now here, I'm going to keep my numerator as is because I know some stuff is going to cancel. I have the limit as x approaches 5, keeping that numerator the same, just factored as is, x minus 5 times the square root of x plus the square root of 5. Now for my denominator, if I fully multiply this out since I am multiplying conjugates by each other, I'm going to end up getting here x minus 5. Now here I see this common factor in my numerator. I have x minus 5. In my denominator, I also have x minus 5. So those totally cancel out here. Now I'm just left to find the limit as x approaches 5 of the square root of x plus the square root of 5. Now from here I can just go ahead and plug 5 in. So doing that, I end up with the square root of 5 plus the square root of 5, which ends up being 2 root 5 fully simplified. So my answer here for the limit of this function as x approaches 5 is 2 root 5. Let me know if you have any questions. Thanks for watching.

1. Limits and Continuity

Finding Limits Algebraically

Business Calculus

1. Limits and Continuity

Finding Limits Algebraically

Business Calculus 때문에 힘들어하고 계신가요?

저희를 신뢰하는 수천 명의 학생들이 시험 합류하세요!

객관식

Find the limit.
$$\lim$_{x$\rarr$5}$\frac{x-5}{\sqrt{x}$-$\sqrt$5}$

$$\sqrt$5$

$2$\sqrt$5$

$\frac{\sqrt5}{10}$

Does not exist

검증된 단계별 안내

Rewrite the given limit expression: $ \lim_{x \to 5} \frac{x - 5}{\sqrt{x} - \sqrt{5}} $. Notice that the denominator contains a square root, which suggests rationalizing the denominator might be helpful.

Multiply the numerator and denominator by the conjugate of the denominator, $ \sqrt{x} + \sqrt{5} $, to eliminate the square root in the denominator. This gives: $ \frac{(x - 5)(\sqrt{x} + \sqrt{5})}{(\sqrt{x} - \sqrt{5})(\sqrt{x} + \sqrt{5})} $.

Simplify the denominator using the difference of squares formula: $ (\sqrt{x} - \sqrt{5})(\sqrt{x} + \sqrt{5}) = x - 5 $. The expression now becomes: $ \frac{(x - 5)(\sqrt{x} + \sqrt{5})}{x - 5} $.

Cancel the common factor $ x - 5 $ in the numerator and denominator (valid as long as $ x \neq 5 $). The simplified expression is: $ \sqrt{x} + \sqrt{5} $.

Finally, evaluate the limit as $ x \to 5 $ by substituting $ x = 5 $ into the simplified expression: $ \sqrt{5} + \sqrt{5} = 2\sqrt{5} $. This is the value of the limit.

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다음 시청

마스터 Finding Limits by Direct Substitution Example 1가 Callie Rethman 짧은 영상으로 설명해 줍니다

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수석 튜터가 만든 문제 세트

전문가 영상 설명

Business Calculus 때문에 힘들어하고 계신가요?

Find the limit.limx→5x−5x−5\(\lim\)_{x\(\rarr\)5}\(\frac{x-5}{\sqrt{x}\)-\(\sqrt\)5}limx→5x−5x−5

다음 시청

Finding Limits Algebraically 연습 세트

Find the limit.
$\(\lim\)_{x\(\rarr\)5}\(\frac{x-5}{\sqrt{x}\)-\(\sqrt\)5}$