Find the limit of 4 / (x^2 * 2^(1/x)) as x approaches 0.

Step-by-step solution

1. 1

   Notice that $1/x$ changes sign across zero. As $x\to 0^{+}$, $1/x\to+\infty$ and $2^{1/x}\to+\infty$; as $x\to 0^{-}$, $1/x\to-\infty$ and $2^{1/x}\to 0^{+}$. Because the factor behaves in opposite ways on the two sides, the one-sided limits must be computed separately — assuming a single answer is the trap here.

2. 2

   Right-hand limit: substitute $t=1/x$. With $x\to 0^{+}$ we have $t\to+\infty$ and $x^2=1/t^2$, so

   $\frac{4}{x^2\cdot 2^{1/x}}=\frac{4}{\frac{1}{t^2}\cdot 2^{t}}=\frac{4t^2}{2^{t}}$

   The substitution converts an awkward $0\cdot\infty$ denominator into a familiar race between a polynomial and an exponential.

3. 3

   Exponential growth wins, so the right-hand limit is 0. For any fixed power, $\displaystyle\lim_{t\to\infty}\frac{t^{n}}{2^{t}}=0$ (two applications of L Hopital settle $n=2$). Hence

   $\lim_{x\to 0^{+}}\frac{4}{x^2\cdot 2^{1/x}}=0$

4. 4

   Left-hand limit: substitute $t=-1/x$. Now $x\to 0^{-}$ gives $t\to+\infty$, $x^2=1/t^2$ and $2^{1/x}=2^{-t}=\dfrac{1}{2^{t}}$, so the expression becomes

   $\frac{4}{\frac{1}{t^2}\cdot\frac{1}{2^{t}}}=4t^2 2^{t}\to+\infty$

   Both factors of the denominator shrink to zero, so the fraction blows up.

5. 5

   Conclude that the two-sided limit does not exist. Since

   $\lim_{x\to 0^{+}}=0\neq +\infty=\lim_{x\to 0^{-}}$

   the two one-sided limits disagree, and a two-sided limit exists only when they agree.

6. 6

   Check the numbers. At $x=0.02$ the value is $8.9\times 10^{-12}$; at $x=-0.02$ it is $1.1\times 10^{19}$. The same expression, evaluated equally close to zero on either side, differs by thirty orders of magnitude — a vivid confirmation that no single limiting value exists.