To evaluate the exponential limits of the form1∞, we use the following result.
Result: if \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=0\) such that \(\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}\) exists, then, \(\underset{x\to a}{\mathop{\lim }}\,{{\left[ 1+f\left( x \right) \right]}^{1/g\left( x \right)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}}}\).
Proof: Let
A \(=\underset{x\to a}{\mathop{\lim }}\,{{\left[ 1+f\left( x \right) \right]}^{\frac{1}{g\left( x \right)}}}\).
⇒ \({{\log }_{e}}A=\underset{x\to a}{\mathop{\lim }}\,\frac{\log \left[ 1+f\left( x \right) \right]}{g\left( x \right)}\).
⇒ \({{\log }_{e}}A=\underset{x\to a}{\mathop{\lim }}\,\frac{\log \left[ 1+f\left( x \right) \right]}{f\left( x \right)}\times \frac{f\left( x \right)}{g\left( x \right)}\).
⇒ \({{\log }_{e}}A=\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}\,\left[ \because \underset{x\to a}{\mathop{\lim }}\,\frac{\log \left[ 1+f\left( x \right) \right]}{f\left( x \right)}=1 \right]\).
\(A={{e}^{\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}}}\).
Remark: the above result can also be restated in the following form:
If \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=1\) and \(\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=\infty \) such that
\(\underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right)-1 \right]g\left( x \right)\) exists.
Then \(\underset{x\to a}{\mathop{\lim }}\,{{\left[ f\left( x \right) \right]}^{g\left( x \right)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right)-1 \right]g\left( x \right)}}\).
Particular cases:
- \(\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{1/x}}=e\).
- \(\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e\).
- \(\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+\lambda x \right)}^{1/x}}={{e}^{\lambda }}\).
- \(\underset{x\to \infty }{\mathop{\lim }}\,{{\left( a+\frac{\lambda }{x} \right)}^{x}}={{e}^{\lambda }}\).
Example: find the polynomial function f (x) of degree 6 satisfying:
\(\underset{x\to 0}{\mathop{\lim }}\,{{\left[ 1+\frac{f\left( x \right)}{{{x}^{3}}} \right]}^{1/x}}={{e}^{2}}\).
Solution: Let f (x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + a₅x⁵ + a₆x⁶.
Then, \(\underset{x\to 0}{\mathop{\lim }}\,{{\left[ 1+\frac{f\left( x \right)}{{{x}^{3}}} \right]}^{1/x}}={{e}^{2}}\).
\({{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{f\left( x \right)}{{{x}^{4}}}}}={{e}^{2}}\).
\(\underset{x\to 0}{\mathop{\lim }}\,\frac{f\left( x \right)}{{{x}^{4}}}=2\).
\(\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+{{a}_{4}}{{x}^{4}}+{{a}_{5}}{{x}^{5}}+{{a}_{6}}{{x}^{6}}}{{{x}^{4}}}=2\).
⇒ a₀ = a₁ = a₂ = a₃ = 0, a₄ = 2
∴ f (x) = 2x⁴ + a₅x⁵ + a₆x⁶, where a₅, a₆ are real numbers.