Question

Evaluate

LIM 3x^2 - (7x+1) sqrt root (x) + 5 .
-------------------------
X---> 1 (X - 1)

Answer 1

do you know l hospital rule?

Answer 2

Were not in there yet just. the basics Rationalizing/Factorization

Answer 3

i really couldnt understand ur numerator..can u please use equation editor to make it readable?

Answer 4

Oh wait sorry

Answer 5

LIM
X->1

Answer 6

3x^2 - (7x+1) square root of (X) +5
-----------------------------------
X-1

Answer 7

\[\sqrt{x+5}??\]

Answer 8

Is this the question?\[\lim_{x \to 1}\frac{3x^2-(7x+1)\sqrt{x}+5}{x-1}\]

Answer 9

Yeah

Answer 10

A bit complicated, forgive me.
\[\lim_{x \to 1}\frac{3x^2-(7x+1)\sqrt{x}+5}{x-1}\]\[=\lim_{x \to 1}\frac{3x^2-(7x+1)\sqrt{x}+5}{(\sqrt{x}-1)(\sqrt{x}+1)}\]\[=\lim_{x \to 1}\frac{3x^2-7x^\frac{3}{2}-\sqrt{x}+5}{(\sqrt{x}-1)(\sqrt{x}+1)}\]To make the substitution works, we need to get \(\sqrt{x}-1\) cancelled. So, we have to factorize the numerator with \(\sqrt{x}-1\) as a factor. Hmm.. Can you do it?

Answer 11

I''ll try But how did it end up in conjugating ? how did it became square root of X in the denominator

Answer 12

\[x-1 = (\sqrt{x})^2 -1 =(\sqrt{x}-1)(\sqrt{x}+1)\]

Answer 13

I think in this case, we don't need conjugate :|

Answer 14

We only need factorization.