Question

If \[ \lim_{x\to 0^+}x^2 \left( \frac{e^{1/x}-e^{-1/x}} {e^{1/x}+e^{-1/x}} \right)=k \] and \[ \lim_{x\to 0^-}x^2 \left( \frac{e^{1/x}-e^{-1/x}} {e^{1/x}+e^{-1/x}} \right)=l, \] then

• A. \(k=l\)
• B. \(k=1,\;l=-1\)
• C. \(k=-1,\;l=1\)
• D. \(k\neq l\neq \pm 1\)

Hint:

When an expression contains \(e^t-e^{-t}\) over \(e^t+e^{-t}\), recognize it as \(\tanh t\). Then check the limiting value of \(t\).

Answer 1

The Correct Option is A

Step 1: Simplify the fraction as tanh(1/x).
\(\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}=\tanh\left(\frac{1}{x}\right)\). As \(x\to0^+\), \(\frac{1}{x}\to+\infty\), so \(\tanh\to1\). Thus \(k=\lim_{x\to0^+}x^2\cdot1=0\).

Step 2: Left-hand limit.
As \(x\to0^-\), \(\frac{1}{x}\to-\infty\), so \(\tanh\to-1\). Thus \(l=\lim_{x\to0^-}x^2\cdot(-1)=0\). Since both \(k=l=0\): \[\boxed{k=l}\]