limit(limit X→oo)(-ln(X) + 2 × X½)\(\lim\limits_{x \to \infty}(-ln

1.	limit(limit X→oo)(-ln(X) + 2 × X½)\(\lim\limits_{x \to \infty}(-ln \, x + 2x\frac 12)\)
Answer» \(\lim\limits_{x \to \infty}(-ln \, x + 2x\frac 12)\) (∞ - ∞ case) Take y = 1/x then limit converts to y → 0 = \(\lim\limits_{y \to 0}(-ln (\frac 1y)+ 2(\frac 1y)^{\frac 12}\) = \(\lim\limits_{y \to 0}(-ln\, 1 - ln\,y) + \frac {2}{y\frac 12}) (∵ ln \frac AB = ln\,A-ln\,B)\) = \(\lim\limits_{y \to 0}(-ln\, y+ \frac {2}{y\frac12})\) (∞ - ∞ case) (∵ ln 1 = 0) = \(\lim\limits_{y \to 0}(\frac {y\frac12\,ln\,y+2}{y\frac 12})\) Now, \(\lim\limits_{y \to 0} y\frac 12 \, ln\,y \) (0 x ∞) = \(\lim\limits_{y \to 0} \frac {ln\,y}{y\frac{-1}{2}}\) (∞/∞ type) = \(\lim\limits_{y \to 0} \frac {\frac 1y}{\frac {-1}{2}y\frac{-3}{2}}\) (By using D.L.H Rule) = \(\lim\limits_{y \to 0} \frac {-2\,y\frac 32}{y}\) = \(\lim\limits_{y \to 0} -2\,y\frac 12\) = -2 x 0 = 0 ∴ \(\lim\limits_{y \to 0} \frac {y\frac 12\, ln\,y+2}{y\frac 12} = \frac {0+2}{0} = \frac 20 = \infty\) (not defined)

limit(limit X→oo)(-ln(X) + 2 × X½)\(\lim\limits_{x \to \infty}(-ln \, x + 2x\frac 12)\)

Answer»

\(\lim\limits_{x \to \infty}(-ln \, x + 2x\frac 12)\) (∞ - ∞ case)

Take y = 1/x then limit converts to y → 0

= \(\lim\limits_{y \to 0}(-ln (\frac 1y)+ 2(\frac 1y)^{\frac 12}\)

= \(\lim\limits_{y \to 0}(-ln\, 1 - ln\,y) + \frac {2}{y\frac 12}) (∵ ln \frac AB = ln\,A-ln\,B)\)

= \(\lim\limits_{y \to 0}(-ln\, y+ \frac {2}{y\frac12})\) (∞ - ∞ case) (∵ ln 1 = 0)

= \(\lim\limits_{y \to 0}(\frac {y\frac12\,ln\,y+2}{y\frac 12})\)

Now, \(\lim\limits_{y \to 0} y\frac 12 \, ln\,y \) (0 x ∞)

= \(\lim\limits_{y \to 0} \frac {ln\,y}{y\frac{-1}{2}}\) (∞/∞ type)

= \(\lim\limits_{y \to 0} \frac {\frac 1y}{\frac {-1}{2}y\frac{-3}{2}}\) (By using D.L.H Rule)

= \(\lim\limits_{y \to 0} \frac {-2\,y\frac 32}{y}\)

= \(\lim\limits_{y \to 0} -2\,y\frac 12\)

= -2 x 0 = 0

∴ \(\lim\limits_{y \to 0} \frac {y\frac 12\, ln\,y+2}{y\frac 12} = \frac {0+2}{0} = \frac 20 = \infty\) (not defined)