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
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
Show 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 \(t\).
Updated On: Jun 26, 2026
- \(k=l\)
- \(k=1,\;l=-1\)
- \(k=-1,\;l=1\)
- \(k\neq l\neq \pm 1\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Simplify the exponential fraction.
We have \[ \frac{e^{1/x}-e^{-1/x}} {e^{1/x}+e^{-1/x}} \] This expression is of the form \[ \tanh\left(\frac{1}{x}\right) \] So, \[ \frac{e^{1/x}-e^{-1/x}} {e^{1/x}+e^{-1/x}} = \tanh\left(\frac{1}{x}\right) \]
Step 2: Evaluate the right hand limit.
As \[ x\to 0^+, \] we get \[ \frac{1}{x}\to +\infty \] Therefore, \[ \tanh\left(\frac{1}{x}\right)\to 1 \] Hence, \[ k=\lim_{x\to 0^+}x^2\cdot 1 \] Since \[ x^2\to 0, \] we get \[ k=0 \]
Step 3: Evaluate the left hand limit.
As \[ x\to 0^-, \] we get \[ \frac{1}{x}\to -\infty \] Therefore, \[ \tanh\left(\frac{1}{x}\right)\to -1 \] Hence, \[ l=\lim_{x\to 0^-}x^2\cdot (-1) \] Since \[ x^2\to 0, \] we get \[ l=0 \]
Step 4: Compare \(k\) and \(l\).
We found \[ k=0 \] and \[ l=0 \] Therefore, \[ k=l \]
Step 5: Final conclusion.
Hence, \[ \boxed{k=l} \]
We have \[ \frac{e^{1/x}-e^{-1/x}} {e^{1/x}+e^{-1/x}} \] This expression is of the form \[ \tanh\left(\frac{1}{x}\right) \] So, \[ \frac{e^{1/x}-e^{-1/x}} {e^{1/x}+e^{-1/x}} = \tanh\left(\frac{1}{x}\right) \]
Step 2: Evaluate the right hand limit.
As \[ x\to 0^+, \] we get \[ \frac{1}{x}\to +\infty \] Therefore, \[ \tanh\left(\frac{1}{x}\right)\to 1 \] Hence, \[ k=\lim_{x\to 0^+}x^2\cdot 1 \] Since \[ x^2\to 0, \] we get \[ k=0 \]
Step 3: Evaluate the left hand limit.
As \[ x\to 0^-, \] we get \[ \frac{1}{x}\to -\infty \] Therefore, \[ \tanh\left(\frac{1}{x}\right)\to -1 \] Hence, \[ l=\lim_{x\to 0^-}x^2\cdot (-1) \] Since \[ x^2\to 0, \] we get \[ l=0 \]
Step 4: Compare \(k\) and \(l\).
We found \[ k=0 \] and \[ l=0 \] Therefore, \[ k=l \]
Step 5: Final conclusion.
Hence, \[ \boxed{k=l} \]
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