Differentiable Limits!!

Calculus Level 2

lim ⁡ x → 1 ( x − 1 ) ( 2 x − 3 ) 2 x 2 + x − 3 = ? \displaystyle\lim_{x\to 1}\dfrac{(\sqrt{x}-1)(2x-3)}{2x^2+x-3} = ?


The answer is -0.1.

Parth Lohomi by Parth Lohomi , 20, India

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2 solutions

Tanishq Varshney
Apr 29, 2015

lim ⁡ x → 1 ( x − 1 ) ( x + 1 ) ( 2 x − 3 ) ( x − 1 ) ( 2 x + 3 ) ( x + 1 ) \displaystyle \lim_{x\to 1} \frac{(\sqrt{x}-1)(\sqrt{x}+1)(2x-3)}{(x-1)(2x+3)(\sqrt{x}+1)}

using property of ( a − b ) ( a + b ) = a 2 − b 2 (a-b)(a+b)=a^2-b^2 in the numerator.

lim ⁡ x → 1 2 x − 3 ( x + 1 ) ( 2 x + 3 ) \displaystyle \lim_{x\to 1} \frac{2x-3}{(\sqrt{x}+1)(2x+3)}

apply the limits

− 1 10 = − 0.1 \frac{-1}{10}=-0.1

6 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice and neat!

User 123 - 6 years, 1 month ago

Very elegantly done!

Aran Pasupathy - 6 years ago

lim ⁡ x → 1 ( x − 1 ) ( 2 x − 3 ) ( x − 1 ) ( 2 x + 3 ) = lim ⁡ x → 1 2 x − 3 ( x + 1 ) ( 2 x + 3 ) = 2 − 3 ( 1 + 1 ) ( 2 + 3 ) = − . 1 \lim_{x\to 1} \dfrac{(\sqrt{x}-1)(2x-3)}{(x-1)(2x+3)} = \lim_{x\to 1} \dfrac{2x-3}{(\sqrt x+1)(2x+3)} = \dfrac{2-3}{(1+1)(2+3)}\\=\boxed{\large -.1}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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