Find the limit.limx→5x−5x−5\lim_{x\rarr5}\frac{x-5}{\sqrt{x}-\sqrt5}limx→5x−5x−5

2 5 2\sqrt5 2 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">

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="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> − 5 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 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.

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

2 5 2\sqrt5 2 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">

5 \sqrt5 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">

5 10 \frac{\sqrt5}{10} 10 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">

1. Limits and Continuity

Finding Limits Algebraically

Business Calculus

1. Limits and Continuity

Finding Limits Algebraically

Struggling with Business Calculus?

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Multiple Choice

Find the limit.
$\lim_{x\rarr5}\frac{x-5}{\sqrt{x}-\sqrt5}$

$\sqrt5$

$2\sqrt5$

$\frac{\sqrt5}{10}$

Does not exist

Verified step by step guidance

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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