# How do you evaluate #[ ( 1 + 3x )^(1/x) ]# as x approaches infinity?

Going to use a nifty wee trick that makes use of the fact that the exponential and natural log functions are inverse operations. This means we can apply both of them without changing the function.

Using the exponent rule of logs we can bring the power down in front giving:

The exponential function is continuous so can write this as

and now just deal with the limit and remember to sub it back into the exponential.

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To evaluate [ ( 1 + 3x )^(1/x) ] as x approaches infinity, we can use the concept of limits.

First, we rewrite the expression as e^(ln(1 + 3x) / x), where e is the base of the natural logarithm.

Next, we take the limit as x approaches infinity.

Using the limit properties, we can simplify the expression further by dividing both the numerator and denominator by x.

This gives us e^(ln(1 + 3x) / x) = e^(ln(1 + 3/x) / 1).

As x approaches infinity, 3/x approaches 0, and ln(1 + 3/x) approaches ln(1) = 0.

Therefore, the expression simplifies to e^0 = 1.

Hence, as x approaches infinity, [ ( 1 + 3x )^(1/x) ] evaluates to 1.

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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.

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.

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.

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