# Thread: Limit of a function as x tends to real number

1. ## Limit of a function as x tends to real number

a) lim ( x tends to 2 ) 1 / ( x - 2 )
b) lim (x tends to 2) 1/ (x-2)^2

I would be grateful if you can help me with this, providing clear explanations

2. Originally Posted by ose90
a) lim ( x tends to 2 ) 1 / ( x - 2 )
b) lim (x tends to 2) 1/ (x-2)^2

I would be grateful if you can help me with this, providing clear explanations

As x tends to 2, x - 2 tends to 0 (obviously), and so will $(x - 2)^2$. In other words, the denominator gets extremely small, and as the denominator gets smaller, $\frac{1}{denominator}$ gets bigger.

So the answer to both is infinity.

3. Good explanation , thanks!

However, the answer shown in my notes for a) is no limit, is it obviously wrong?

4. Originally Posted by ose90
Good explanation , thanks!

However, the answer shown in my notes for a) is no limit, is it obviously wrong?
Yes.
Consider x to 2, with x>2 (from the right). Then x-2>0.
So it will be positive infinity.

Consider x to 2, with x<2 (from the left). Then x-2<0.
So it will be negative infinity.

Thus $\lim_{\substack{x \to 2 \\ x>2}} \quad \frac{1}{x-2} \neq \lim_{\substack{x \to 2 \\ x<2}} \quad \frac{1}{x-2}$

This means the limit doesn't exist, because the limit is not the same from the right and from the left.

5. Originally Posted by Prove It
As x tends to 2, x - 2 tends to 0 (obviously), and so will $(x - 2)^2$. In other words, the denominator gets extremely small, and as the denominator gets smaller, $\frac{1}{denominator}$ gets bigger.

So the answer to both is infinity.
It's essential that left hand and right and right hand limits are always considered.

It's not good enough to say x - 2 --> 0 etc.

You need to consider $x \rightarrow 2^+$ and $x \rightarrow 2^-$, whether the limiting value of x - 2 etc. is $0^+$ or $0^-$ and consequently whether the left and right handed limits are +oo or -oo ....

6. Thanks everyone, I have understood.