Math, asked by kritikkritik691, 1 month ago

lim x tends to 0 sin x+ax/x(x+1)​

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Answered by mathdude500
6

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:\displaystyle\lim_{x \to 0}  \frac{sinx + ax}{x(x + 1)}

If we substitute directly x = 0, we get

\rm \:  =  \:   \dfrac{sin0 + 0}{0(0+ 1)}

\rm \:  =  \:   \dfrac{0}{0}

which is indeterminant form.

Consider again,

\rm :\longmapsto\:\displaystyle\lim_{x \to 0}  \frac{sinx + ax}{x(x + 1)}

\rm \:  =  \: \displaystyle\lim_{x \to 0}  \frac{1}{x + 1}  \times \displaystyle\lim_{x \to 0}  \frac{sinx + ax}{x}

\rm \:  =  \: \dfrac{1}{1 + 0} \times \displaystyle\lim_{x \to 0} \bigg[\dfrac{sinx}{x}  + \dfrac{ax}{x} \bigg]

\rm \:  =  \: 1 \times  \bigg[\displaystyle\lim_{x \to 0} \dfrac{sinx}{x}  +\displaystyle\lim_{x \to 0}  \dfrac{ax}{x} \bigg]

\rm \:  =  \: 1 + a

Thus,

\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}  \frac{sinx + ax}{x(x + 1)}  = a + 1 \: }}

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Learn More :-

\boxed{\tt{ \displaystyle\lim_{x \to 0}  \:  \frac{sinx}{x} = 1 \: }}

\boxed{\tt{ \displaystyle\lim_{x \to 0}  \:  \frac{tanx}{x} = 1 \: }}

\boxed{\tt{ \displaystyle\lim_{x \to 0}  \:  \frac{log(1 + x)}{x} = 1 \: }}

\boxed{\tt{ \displaystyle\lim_{x \to 0}  \:  \frac{ {e}^{x}  \:  -  \: 1}{x} = 1 \: }}

\boxed{\tt{ \displaystyle\lim_{x \to 0}  \:  \frac{ {a}^{x}  \:  -  \: 1}{x} = loga \: }}

Answered by ProximaNova
5

Step-by-step explanation:

\lim \limits_{x \to 0} \dfrac{sinx + ax}{x^2+x}

Putting x = 0,

\sf \bf :\longmapsto \lim \limits_{x \to 0} \dfrac{sinx + ax}{x^2+x} = \dfrac{0}{0}

So, applying L hospital rule,

\sf \bf :\longmapsto \lim \limits_{x \to 0} \dfrac{cosx + a}{2x+1}

Now putting x = 0,

\sf \bf :\longmapsto \lim \limits_{x \to 0} \dfrac{sinx + ax}{x^2+x} = \dfrac{cos0 + a}{2(0)+1}

\sf \bf :\longmapsto \lim \limits_{x \to 0} \dfrac{sinx + ax}{x^2+x} = \dfrac{a+1}{1}

\sf \bf :\longmapsto \lim \limits_{x \to 0} \dfrac{sinx + ax}{x^2+x} = a+1

As, cos0 = 1

Hence ,

\boxed{\sf \bf \lim \limits_{x \to 0} \dfrac{sinx + ax}{x^2+x} = a+1}

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