Question

lim x tends to 0 sin x+ax/x(x+1)​

Answer 1

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

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

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 :-

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

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

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

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

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

Answer 2

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 ,

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