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$f(x)=\frac{2x^2+|x|}{x}$, show that $\lim_{x\to 0^-} f(x)=-1$

My solution:
$x < 0$
Step 1: Therefore $y=(2x^2-x)/x$
Step 2: $y=2x-1$
Step 3: As $x$ approaches $0$ from left, the difference between $0$ and $x$ diminishes $\implies$ $y$ approaches $-1$ from left and the difference between $-1$ and $y$ diminishes
Therefore $$|y-(-1)|=1-y=1-(2x-1)=2-2x=2(1-x)$$ Step 4: Let $2(1-x) < ε$
$\implies 1-x < ε/2$
$\implies x>1-ε/2$
I have to prove that $ε/2<x<0$

Where did I go wrong?

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3 Answers 3

up vote 2 down vote accepted

There is a little minus sign glitch in the calculation. In the displayed line, we should have $$|y-(-1)|=|y+1|=|(2x-1)+1|=|2x|.$$ Now it is easy to see that if $|x|<\epsilon/2$, and $x$ is negative, then $|y-(-1)|<\epsilon$. (We need to have $x$ negative so that the expression $2x-1$ for $y$ will be correct.)

Comment: I assume that you are expected to use "$\epsilon$-$\delta$." But at the informal level, it is clear that $\lim_{x\to 0^-} (2x-1)=-1$.

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thanx for the explanation, but I am still confused, when I know that y is negative and I wish to get rid of the modulus sign, I put a negative sign in front of y, but it leads to something else as in steps done by me!!! –  Vikram Oct 9 '11 at 18:07
    
@Vikram: You correctly handled $|x|$, when $x$ is negative, and got that $y=2x-1$. And indeed $y$ is negative. But we are not interested in the absolute value of $y$, as you noted. We want the absolute value of $y-(-1)$. This is the absolute value of $2x$ (see my calculation). It so happens that $x$ is negative, so we will want $-\epsilon<2x<0$, but I expressed this using absolute values. –  André Nicolas Oct 9 '11 at 19:31

When $x<0$ then $|x|=-x$. Therefore one has for all $x<0$ the equality $$f(x)={2x^2-x\over x}=2x-1\ ,$$ and as the right side is continuous at $x=0$ it follows that $$\lim_{x\to0-}f(x)=(2x-1)\bigr|_{x=0}=-1\ .$$

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My solution:

First of all devide numerator and denominator by $x$

$$\lim_{x \to 0^-} \frac{2x+\frac{|x|}{x}}{\frac{x}{x}}$$

Observe that we are interested only in case when $|x|=-x$ so we may write:

$$\lim_{x \to 0^-} \frac{2x+\frac{|x|}{x}}{\frac{x}{x}}=\lim_{x \to 0^-} \frac{o+\frac{-x}{x}}{1}=\lim_{x \to 0^-}\frac{0-1}{1}=-1$$

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1  
thanx pedja, but I am supposed to solve this problem with ϵ-δ –  Vikram Oct 9 '11 at 18:08

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