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How do you find #lim 1+1/x# as #x->0^-#?

Answer 1

#lim_(xrarr0^-) = -oo#

You're trying to find the limit of the equation as #x# approaches #0# from the left.

One way to do this is to substitute values in for #x# that are successively closer and closer to #0# (negative values, since it's from the left), until the limit becomes understood:

#1+1/(-0.1) = -9#

#1+ 1/(-0.01) = -99#

#1 + 1/(-0.001) = -999#

The limit will intuitively keep growing negatively, so the limit as #x# approaches #0# will be #-oo#.

Ultimately, if you have the option to, graphing the equation is a surefire way to check this limit: graph{1+1/x [-5, 5, -7.8, 7.8]}

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Answer 2

Emily Ammerman

The limit of 1 + 1/x as x approaches 0 from the negative side is negative infinity.

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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.

How do you find #lim 1+1/x# as #x-&gt;0^-#?

How do you find #lim 1+1/x# as #x->0^-#?