Question

सीमाओं के मान प्राप्त कीजिए : $\lim_{x\rightarrow0}\dfrac{\sin ax}{bx}$

Answer 1

Answer:

$\large\boxed{\sf{\dfrac{a}{b}}}$

Step-by-step explanation:

$\lim_{x\rightarrow0} \frac{ \sin(ax) }{ \sin(bx) } \\ \\ =\lim_{x\rightarrow0} \frac{ \sin(ax) }{ax} \times ax \times \frac{1}{ \frac{ \sin(bx) }{bx} \times bx} \\ \\ = 1 \times ax \times \frac{1}{bx} \\ \\ = \frac{a}{b}$

Concept Map :-

• $\lim_{x\rightarrow0} \dfrac{\sin x}{x}= 1$

Answer 2

Question:-

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Evaluate $\displaystyle\sf {\lim_{x\to0}\dfrac {\sin (ax)}{bx}.}$

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Answer:-

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$\displaystyle\large\boxed {\sf {\lim_{x\to0}\dfrac {\sin (ax)}{bx}=\dfrac {a}{b}}}$

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Solution:-

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We know that,

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• $\displaystyle\sf {\lim_{x\to0}\dfrac {\sin x}{x}=1\quad\quad\dots(1)}$

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Then,

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$\displaystyle\longrightarrow\sf {\lim_{x\to 0}\dfrac {\sin (ax)}{bx}=\lim_{x\to 0}\dfrac {a\sin (ax)}{abx}}$

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$\displaystyle\longrightarrow\sf {\lim_{x\to 0}\dfrac {\sin (ax)}{bx}=\lim_{x\to 0}\dfrac {a\sin (ax)}{b\cdot ax}}$

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$\displaystyle\longrightarrow\sf {\lim_{x\to 0}\dfrac {\sin (ax)}{bx}=\dfrac {a}{b}\lim_{x\to 0}\dfrac {\sin (ax)}{ax}\quad\quad\dots(2)}$

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Well, as $\displaystyle\sf {x\longrightarrow0,\ ax\longrightarrow 0.}$

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Then (2) becomes,

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$\displaystyle\longrightarrow\sf {\lim_{x\to 0}\dfrac {\sin (ax)}{bx}=\dfrac {a}{b}\lim_{ax\to 0}\dfrac {\sin (ax)}{ax}}$

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By (1),

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$\displaystyle\longrightarrow\sf {\lim_{x\to 0}\dfrac {\sin (ax)}{bx}=\dfrac {a}{b}\cdot1}$

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$\displaystyle\longrightarrow\sf {\underline {\underline {\lim_{x\to 0}\dfrac {\sin (ax)}{bx}=\dfrac {a}{b}}}}$