The value of limx - 0-0x2sec2 t dt-x sin x is

X →0lim∫_0^x^2sec^2 tdt/x sin x=x →0lim∫_0^x^2sec^2 tdt/x^2(sinx/x ) =x →0lim∫_0^x^2sec^2 tdt/x^2 (∵x →0limsinx/x=1) Applying L Hospital's rule =x →0lim2xsec^2x ...

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Byju's Answer

Standard XIII

Mathematics

Property 5

The value of ...

Question

The value of $l i m_{x \to 0} \frac{\int_{0}^{x^{2}} s e c^{2} t d t}{x s i n x}$ is

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Solution

The correct option is C 1
$l i m x \to 0 \frac{\int_{0}^{x^{2}} s e c^{2} t d t}{x s i n x} = l i m x \to 0 \frac{\int_{0}^{x^{2}} s e c^{2} t d t}{x^{2} (\frac{s i n x}{x})}$
$= l i m x \to 0 \frac{\int_{0}^{x^{2}} s e c^{2} t d t}{x^{2}}$
( $∵ l i m x \to 0 \frac{s i n x}{x} = 1$ )
$A p p l y i n g L H o s p i t a l^{'} s r u l e$
$= l i m x \to 0 \frac{2 x s e c^{2} x}{2 x} = 1$

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The value of limx→0∫x20sec2t dtx sin x is

The correct option is C 1limx→0∫x20sec2tdtx sin x=limx→0∫x20sec2tdtx2(sinxx) =limx→0∫x20sec2tdtx2 (∵limx→0sinxx=1) Applying L Hospital′s rule =limx→02xsec2x2x=1

The value of $l i m_{x \to 0} \frac{\int_{0}^{x^{2}} s e c^{2} t d t}{x s i n x}$ is