Let p-limx- 0- 1-tan2-x1-2x then log p is equal to-

Ln P=lim_x→ 0^+1/2xln(1+tan^2√(x)) lim_x→ 0^+1/xln(sec√(x)) Applying L hospital's rule: =lim_x→ 0^+sec√(x)tan√(x)/sec√(x).2√(x) =lim_x→ 0^+tan√(x)/2√(x) =1/2 ...

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Byju's Answer

Standard XII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

Let p=limx→ 0...

Question

Let $p = {lim}_{x \to 0^{+}} (1 + t a n^{2} \sqrt{x})^{\frac{1}{2 x}}$ then log p is equal to:

\frac{1}{2}

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\frac{1}{4}

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Solution

The correct option is A $\frac{1}{2}$
$l n P = {lim}_{x \to 0^{+}} \frac{1}{2 x} l n (1 + t a n^{2} \sqrt{x}) {lim}_{x \to 0^{+}} \frac{1}{x} l n (s e c \sqrt{x})$
Applying L hospital's rule:
$= {lim}_{x \to 0^{+}} \frac{s e c \sqrt{x} t a n \sqrt{x}}{s e c \sqrt{x} .2 \sqrt{x}} = {lim}_{x \to 0^{+}} \frac{t a n \sqrt{x}}{2 \sqrt{x}} = \frac{1}{2}$

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