Evaluation of Limits of the Form 1∞

Evaluation of Limits of the Form 1

 To evaluate the exponential limits of the form 1, 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:

a) \(\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{1/x}}=e\).

b) \(\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e\).

c) \(\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+\lambda x \right)}^{1/x}}={{e}^{\lambda }}\).

d) \(\underset{x\to \infty }{\mathop{\lim }}\,{{\left( a+\frac{\lambda }{x} \right)}^{x}}={{e}^{\lambda }}\).

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.