Question

Limit x-->1 sqrt x^2 +3x-2 /x-1 find the limit if exist

Answer 1

Does square root cover only the numerator or the entire fraction?

Answer 2

Also, what level maths is this? Are you allowed to use L'Hospitals rule?

Answer 3

we can use L'hospital but we didnt learn it yet

Answer 4

\[\lim -> 1 \sqrt{x^2+3x}-2/x-1\]

Answer 5

i just kind of stuck so far what i got is (x^2+3x)-4/x-1*Sqrt (x^2+3x+2)

Answer 6

Well, lhospital makes it very simple. As long as the derivates arent zero you can diferenciate the numerator and the denominator. The resulting fraction will have the same limit as the original.

In this case it isnt necessary however:
\[\lim_{x \rightarrow 1} { \sqrt{x^2 + 3x } - 2 \over x-1} = { \sqrt{x^2 + 3x } - 2 \over x-1} { \sqrt{x^2 + 3x } + 2 \over\sqrt{x^2 + 3x } + 2} = {{ x^2 + 3x } - 4 \over (x-1)(\sqrt{x^2 + 3x } + 2) }\]
\[{{ x^2 + 3x } - 4 \over (x-1)(\sqrt{x^2 + 3x } + 2) } = { (x-1)(x+4) \over (x-1)(\sqrt{x^2 + 3x } + 2) } = { (x+4) \over(\sqrt{x^2 + 3x } + 2) } = { 5 \over 4 }\]

Answer 7

thanks i got the same result as you soati and i already found the solution