# lim (X2 - 1/X) X->0-i am taking the limits and vertical asymptotes and i cant really get them so i have several questions but i asked only one of them . so can you help me please as soon as...

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### 2 Answers

Hi,

Just to share. To key x to the power of 2, usually we just key x^2 .

For your question:

### lim (X^2 - 1/X) X->0-

Do you mean

Case 1: (x^2 - 1)/x

or

Case 2: (x^2 - 1/x) ?

I have done both cases for you:

Case 1

Take a peek at what the graph (the green one) is like by clicking on the first link I had provided at the bottom to appreciate it.

The expression is: (x^2 - 1)/x

For very small values of x (0<x<1), x^2 << 1

Hence the function approximates to -1/x .

With small negative values of x (0-), the expression becomes +infinity

**Lim (x^2 -1)/x , x->0- = +infinity**

___

Case 2: (x^2 - 1/x)

For very small values of x (0<x<1), x^2 becomes practically negligible and the term -1/x dominates.

Again, the expression approximates to just -1/x

**Lim (x^2 - 1/x) , x->0- = +infinity**

CONCLUSION

In either case, the answer is still +infinity

(Apologies any confusion caused prior to this ammendment on 3/12)

Click on the links below, whichever the case you intended, and appreciate the question. It's fun! :)

To find the Lt(x^2-1/x)/(x) as x--> 0

The expression (x^2-1/x)/x = (x^3-1)/x

We know that a^3-1 = (a-1)(a^2+a+1).

Therefore (x^2-1/x)/x = (x^3-1)/x = (x-1)(x^2+x+1/x)/x

(x^3-1)/x = (x-1)(x^2+x+1/x)/x.

(x^2-1/x)/x = (x-1){x+1-1/x^2).

Therefore Lt {(x-1){x+1+1/x^2) as x--> 0- } = (0 -1) {0+1+1/ (-e)^2} , where 0 < e < any small number.

Lt { (x-1)(x+ 1+1/x^2) as x-->0-} = -1/(-e)^2 as e--> 0+ is - infinity.

Therefore the limit { (x^2-1/x)/x as x--> 0-} = -infinity.