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What is #lim_ (xrarr0) (1/x -1/x^2-x)#?

Answer 1

Luke Alvarez

Since both limits from left and right are #-infty#, the limit is also #-infty#.

Since #x->0#, the last term #-x# is irrelevant. So, let's focus on the rest.

We have that, if #x->0^-#, then #1/x -> -infty#, and so does #-1/x^2#. So we have #-infty-infty = -infty#.

On the other hand, if #x->0^+#, then we are in a #infty-infty# form, which we can't solve directly.

Thus, we have to add up the two fractions:

#1/x-1/x^2 = (x^2-x)/x^3 = (x-1)/x^2#

And now the limit as #x->0^+# of this quantity is #(-1)/0^+ = -infty#,

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Answer 2

Paisley Johnson

The limit of (1/x - 1/x^2 - x) as x approaches 0 is 0.

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Answer 3

Emily Ammerman

To find ( \lim_{x \to 0} \left(\frac{1}{x} - \frac{1}{x^2 - x}\right) ), you can simplify the expression first by finding a common denominator for the two fractions, then combine them into a single fraction. Once simplified, you can directly substitute ( x = 0 ) into the expression to evaluate the limit.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.