Limits of Function II

Calculus Level 2

Find the limit

lim ⁡ x → 1 ( x − 1 x 2 + 3 − 2 ) . \large \lim _{x\to1 }\left(\frac{x-1}{\sqrt{x^2+3}-2}\right).


The answer is 2.

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Dimitris Kouroupos by dimitris kouroupos , 23, Greece

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2 solutions

Chew-Seong Cheong
Feb 13, 2017

L = lim ⁡ x → 1 x − 1 x 2 + 3 − 2 A 0/0 cases, L’H o ˆ pital’s rule applies. = lim ⁡ x → 1 1 1 2 ⋅ 2 x x 2 + 3 Differentiate up and down w.r.t. x . = 1 1 2 = 2 \begin{aligned} L & = \lim_{x \to 1} \frac {x-1}{\sqrt{x^2+3}-2} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 1} \frac 1{\frac 12 \cdot \frac {2x}{\sqrt{x^2+3}}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x. \\ & = \frac 1{\frac 12} = \boxed{2} \end{aligned}

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Relevant wiki: Limits by conjugates

We observe that the limit results in undefined 0 0 \frac{0}{0} ,Next we must change the limit

lim ⁡ x → 1 ( x − 1 x 2 + 3 − 2 ) = lim ⁡ x → 1 ( ( x − 1 ) ( x 2 + 3 + 2 ) ( x 2 + 3 − 2 ) ( x 2 + 3 + 2 ) ) = lim ⁡ x → 1 ( ( x − 1 ) ( x 2 + 3 + 2 ) x 2 − 1 ) = lim ⁡ x → 1 ( ( x 2 + 3 + 2 ) ( x + 1 ) ) = 2 \large \lim _{x\to1 }\left(\frac{x-1}{\sqrt{x^2+3}-2}\right)= \ \lim _{x\to1 }\left(\frac{(x-1)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}\right)= \lim _{x\to1 }\left(\frac{(x-1)(\sqrt{x^2+3}+2)}{x^2-1}\right)= \lim _{x\to1 }\left(\frac{(\sqrt{x^2+3}+2)}{(x+1)}\right)=2

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See bro, In 0/0 form we can use L'Hopitals rule

So applying it we get

lim x--> 1 x 2 + 3 x \dfrac{\sqrt{x^2+3}}{x}

So Answer is 2 \boxed{2}

Md Zuhair - 4 years, 3 months ago

Haha You did the full sum correctly but inserted the limit value wrong.

See LAst line lim ⁡ x → 1 ( ( x 2 + 3 + 2 ) ( x + 1 ) ) = 2 + 2 2 = 4 2 = 2 \lim _{x\to1 }\left(\frac{(\sqrt{x^2+3}+2)}{(x+1)}\right)=\dfrac{2+2}{2} = \dfrac{4}{2} = \boxed{2}

Md Zuhair - 4 years, 3 months ago

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Thanks. I've updated the answer to 2. Those who previously answered 2 has been marked correct.

Brilliant Mathematics Staff - 4 years, 3 months ago

I would like to apologize for my mistake ,thank you for a comment

dimitris kouroupos - 4 years, 3 months ago

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No no. Its ok. We just got it correct.

Md Zuhair - 4 years, 3 months ago

Sorry for my mistake

dimitris kouroupos - 4 years, 3 months ago

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