AP Calculus AB and BC

Limit of e^(1/x) as x Approaches 0 from the Left

Q: Why does the answer depend so much on the side?

Because 1/x changes sign across zero, and the exponential treats large positive and large negative exponents completely differently: one explodes, the other collapses.

The limit of e to the one over x as x approaches zero from the left is zero. On the left the exponent one over x runs to negative infinity, and the exponential of a large negative number collapses to zero.

\lim_{x \to 0^-} e^{1/x} = 0

Settled by tracking the exponent, then end behaviour of the exponential.

Follow the exponent

Work inside out. As $x \to 0^{-}$ , the exponent $1/x \to -\infty$ , because a small negative denominator gives a large negative quotient.

\lim_{x \to 0^{-}} e^{1/x} = \lim_{u \to -\infty} e^{u} = 0

The substitution $u = 1/x$ makes it a standard end-behaviour question about the exponential, and $e^{u} \to 0$ as $u \to -\infty$ .

The other side is completely different

As $x \to 0^{+}$ the exponent runs to $+\infty$ instead, so $e^{1/x} \to +\infty$ . One side gives 0, the other gives unbounded growth, so the two-sided limit does not exist.

This function is the standard example of a discontinuity that is neither removable nor a jump. Both one-sided behaviours are perfectly well understood, and they simply have nothing to do with each other.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why does the answer depend so much on the side?

Because $1/x$ changes sign across zero, and the exponential treats large positive and large negative exponents completely differently: one explodes, the other collapses.

Is this a jump discontinuity?

No. A jump needs both one-sided limits to exist and be finite. Here the right-hand limit is infinite, so this is an infinite discontinuity.