Evaluation of limits of the form1∞

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.