How can I solve a one-sided limit without using l'Hopital's rule?

[squirrelschaser]

Homework Statement

Find the $lim _{x-> -1+} sqrt(x^2-3x)-2/|x+1|$

Homework Equations

The Attempt at a Solution

I can only solve it using l'hopital rule and would like to know the steps of solving it without using it.

$lim _{x->-1+} (2x-3)/|1|= -5/4$

 

[Mark44]

Homework Statement

Find the $lim _{x-> -1+} sqrt(x^2-3x)-2/|x+1|$

Homework Equations

The Attempt at a Solution

I can only solve it using l'hopital rule and would like to know the steps of solving it without using it.

$lim _{x->-1+} (2x-3)/|1|= -5/4$

Multiply the expression by 1 in the form of the conjugate of the numerator over itself. The |x + 1| factor in the denominator can be replaced by x + 1, since x is to the right of -1, so x + 1 > 0. If the limit had been as x approaches -1 from the left you have to replace |x + 1| by -(x + 1).

 

[LCKurtz]

Homework Statement

Find the $lim _{x-> -1+} sqrt(x^2-3x)-2/|x+1|$

Homework Equations

The Attempt at a Solution

I can only solve it using l'hopital rule and would like to know the steps of solving it without using it.

$lim _{x->-1+} (2x-3)/|1|= -5/4$

I suppose you mean$$
\lim_{x\to -1^+}\frac{\sqrt{x^2-3x}-2}{|x+1|}$$which is not what you wrote. Anyway since $x>-1$ you can write $|x+1|=x+1$. Try rationalizing the numerator and see if you can get it then.

[Edit] Mark44 must type faster than I do.

 

[squirrelschaser]

I'm dumb. Much thanks.

 

[HallsofIvy]

No, you are careless- that, at least, is curable!