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How do you find the limit of #(2+x)^(1/x)# as x approaches #0^-#?

Answer 1

The limit is #0#. For "how" please see below.

Informally

As #xrarr0-#, #(2+x)rarr2# and #1/x rarr-oo#.

So as #xrarr0^-# we have a quantity close to #2# being raised to a very big negative power.

#2# to a big negative is the same as #1# over (#2# to a big positive power) which is #1# over a big number and that is close to #0#.

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Answer 2

Aurora Alvarado

To find the limit of (2+x)^(1/x) as x approaches 0^-, we can use the concept of exponential limits.

First, we rewrite the expression as e^(ln(2+x) / x).

Next, we take the limit of the exponent as x approaches 0^-.

Using the limit properties, we have ln(2+x) / x = ln(2+x) / (x - 0) = ln(2+x) / x.

Now, we can evaluate the limit of ln(2+x) / x as x approaches 0^-.

By applying L'Hôpital's rule, we differentiate the numerator and denominator with respect to x.

The derivative of ln(2+x) is 1 / (2+x), and the derivative of x is 1.

Taking the limit again, we have 1 / (2+0) = 1/2.

Therefore, the limit of (2+x)^(1/x) as x approaches 0^- is e^(1/2).

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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.