lim x->0+ for 1/1+ (5^(1/x))
I think there's no limit, since there is a vertical tangent at 0, is this correct?
To find the limit of the function as x approaches 0 from the positive side, we can first simplify the expression:
1 / (1 + 5^(1/x))
Since we are approaching 0 from the positive side, let's evaluate the expression as x gets closer to 0.
As x approaches 0, 5^(1/x) approaches 5^∞, which is equivalent to 5 raised to an infinitely large power. To determine the behavior of this expression, let's analyze the exponent.
As x approaches 0 from the positive side, 1/x becomes infinitely large. This means that 5^(1/x) becomes 5^∞, which can be interpreted as an indeterminate form.
Now, let's focus on the denominator, 1 + 5^(1/x). As x approaches 0 from the positive side, 5^(1/x) approaches infinity, making the denominator tend toward infinity as well.
When the denominator approaches infinity, the overall fraction tends towards 0. Therefore, we can conclude that the limit as x approaches 0 from the positive side for the given function is 0.
Regarding the mention of a vertical tangent at 0, it's important to note that this concept applies to derivatives in the context of calculus. The presence or absence of a vertical tangent does not directly determine the existence or value of a limit.
Thus, in this case, the limit is well-defined and equals 0.