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

Answer 1

Isabella Ackerman

#oo#

We may split the limit up as follows, recalling the fact that

#lim_(x->a)[f(x)+-g(x)]=lim_(x->a)f(x)+-lim_(x->a)g(x)#

Then,

#lim_(x->0^+)(1+1/x)=lim_(x->0^+)1+lim_(x->0^+)1/x#

#lim_(x->0^+)1=1,# in general, the limit to any value of a constant is simply that constant.

To determine #lim_(x->0^+)1/x,# envision dividing #1/x# by smaller and smaller positive numbers, as we're approaching #0# from the positive side:

#1/0.1=10# #1/0.01=100# #1/0.001=1000# #1/0.0001=10000#

We can see as we approach #0# from the positive side, our result gets larger and larger, that is, heads toward #oo.#

Thus, #lim_(x->0^+)1/x=oo#

Our result is

#lim_(x->0^+)(1+1/x)=1+oo=oo,# the #1# doesn't change the fact that we're heading toward infinity, it doesn't impact the answer, #1# is insignificant in comparison.

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Answer 2

Ezra Adams

The limit of 1 + 1/x as x approaches 0 from the positive side (x->0+) is equal to 1.

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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^+#?