Limits – I
Limits of The Form \(\underset{x\to a}{\mathop{\lim }}\,{{\left( f(x) \right)}^{g(x)}}\).
1. Form 0⁰, ∞⁰.
Let \(L=\underset{x\to a}{\mathop{\lim }}\,{{\left( f(x) \right)}^{g(x)}}\) then
\({{\log }_{e}}L={{\log }_{e}}\left[ \underset{x\to a}{\mathop{\lim }}\,{{\{f(x)\}}^{g(x)}} \right]\).
\({{\log }_{e}}L=\underset{x\to a}{\mathop{\lim }}\,\left[ {{\log }_{e}}{{\{f(x)\}}^{g(x)}} \right]\).
\({{\log }_{e}}L=\underset{x\to a}{\mathop{\lim }}\,g(x){{\log }_{e}}\left[ f(x) \right]\).
\(L={{e}^{\underset{x\to a}{\mathop{\lim }}\,g(x)\left[ {{\log }_{e}}(f(x) \right])}}\).
Example 1: Evaluate \(\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{1/x}}\).
Solution: Given that \(\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{1/x}}={{e}^{\log \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{\frac{1}{x}}}}}\).
\(={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\log {{x}^{\frac{1}{x}}}}}\).
\(={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{x}\log x}}\).
(since x increases faster than loge x when x → ∞)
= e⁰
= 0
Example 2: Evaluate \(\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}\).
Solution: Given that \(\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}\).
Let \(y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}\).
\(\log y=\log \underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}\).
\(\log y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\log {{(\cos x)}^{\cos x}}\).
\(\log y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\cos x.\log (\cos x)\).
\(\log y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\frac{1}{\sec x}.\log (\cos x)\).
Using L – Hospital’s rule
\(=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\frac{\frac{-\sin x}{\cos x}}{\sec x.\tan x}\).
\(=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,(-\cos x)\).
= 0
logy = 0
y = e⁰
y = 1