# [(Limit x->0- ) = - infinity] help w this question

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#1
So this just happened in our class. I don't know anything about limits, no clue what's going on. Can someone please explain what's going on here? Where did the LHS, RHS come from? Why is LHS = negative infinity?

For reference I just finished A lvl maths. Sorry the handwriting is so bad, I'll try typing it here:

Q. Let y = 1/x and find limx->0 1/x = ??

Ans. LHS= limx->0- 1/x = - infinity (*negative infinity)
RHS = limx->0+ 1/x = infinity

therefore, LHS is not equal to RHS
limit doesn't exist

Last edited by newstudent1234; 3 weeks ago
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3 weeks ago
#2
The idea of the LHS is: what happens to the value of 1/x as x gets closer and closer to zero, approaching zero from negative values of x? (It is the superscript of - that means "approaching from below", or from the left of zero on a number line.) By trying some values of x such as -1, -0.1, -0.01, etc, it should be clear that the value of 1/x is getting larger and larger, but is always negative. So this value is getting closer to negative infinity.

The RHS is the same, but in this case you are approaching zero from above, giving you a limit of positive infinity.

As the two results are different, there is no such thing as the limit as x gets closer to zero of 1/x. It depends how you approach zero, and if there was such a limit, this wouldn't matter.

It's worth noting that some texts would say that both of these limits are undefined, reserving the idea of a limit as a finite number. That wouldn't stop the above argument from working, but a bit of different notation would be required.
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#3
(Original post by Pangol)
The idea of the LHS is: what happens to the value of 1/x as x gets closer and closer to zero, approaching zero from negative values of x? (It is the superscript of - that means "approaching from below", or from the left of zero on a number line.) By trying some values of x such as -1, -0.1, -0.01, etc, it should be clear that the value of 1/x is getting larger and larger, but is always negative. So this value is getting closer to negative infinity.

The RHS is the same, but in this case you are approaching zero from above, giving you a limit of positive infinity.

As the two results are different, there is no such thing as the limit as x gets closer to zero of 1/x. It depends how you approach zero, and if there was such a limit, this wouldn't matter.

It's worth noting that some texts would say that both of these limits are undefined, reserving the idea of a limit as a finite number. That wouldn't stop the above argument from working, but a bit of different notation would be required.
Thank you so much for your help! The way you explained was so easy to understand. Super super thanks!!
0
3 weeks ago
#4
lim
x->a 1/x-a as x----> a 1/x---- > infinity 8
0
X

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