Thread: 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

Thanks in advance!

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

Thanks in advance!

As x tends to 2, x - 2 tends to 0 (obviously), and so will . In other words, the denominator gets extremely small, and as the denominator gets smaller, gets bigger.

So the answer to both is infinity.

Good explanation , thanks!

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

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

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

Originally Posted by Prove It

As x tends to 2, x - 2 tends to 0 (obviously), and so will . In other words, the denominator gets extremely small, and as the denominator gets smaller, 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 and , whether the limiting value of x - 2 etc. is or and consequently whether the left and right handed limits are +oo or -oo ....

Thanks everyone, I have understood.