Question

find the limit from attached file

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Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 0} \dfrac{sin(2 + x) - sin(2 - x)}{x}$

As we knw that,

First we put the value of x = 0 in equation, and check their indeterminant form, we get.

$\sf \implies \lim_{x \to 0} \dfrac{sin(2 + 0) - sin(2 - 0)}{0}$

$\sf \implies \lim_{x \to 0} \dfrac{sin(2) - sin(2)}{0}$

$\sf \implies \lim_{x \to 0} \dfrac{0}{0}$

As we can see that,

It is the form of 0/0 indeterminant form,

We can simply apply L HOSPITAL'S RULE.

$\sf \implies \lim_{x \to 0} \dfrac{sin(2 + x) - sin(2 - x)}{x}$

Differentiate w.r.t x, we get.

⇒ sin(2 + x).

⇒ d(sin(2 + x)/dx

⇒ cos(2 + x).1.

⇒ cos(2 + x).

Differentiate w.r.t x, we get.

⇒ sin(2 - x).

⇒ d(sin(2 - x))/dx.

⇒ cos(2 - x)(-1).

⇒ -cos(2 - x).

We can write equation as,

$\sf \implies \lim_{x \to 0} \dfrac{\dfrac{d(sin(2 + x))}{dx}\ \ - \dfrac{d(sin(2 - x))}{dx} }{\dfrac{d(x)}{dx} }$

$\sf \implies \lim_{x \to 0} \dfrac{cos(2 + x) - (-cos(2 - x))}{1}$

$\sf \implies \lim_{x \to 0} = {cos(2 + x) + cos(2 - x)}$

Put the value of x = 0 in equation, we get.

$\sf \implies \lim_{x \to 0} = cos(2 + 0) + cos(2 - 0)$

$\sf \implies \lim_{x \to 0} = cos(2) + cos(2)$

$\sf \implies \lim_{x \to 0} = 2cos(2)$

                                                                                                                 

MORE INFORMATION.

NOTE:

If function takes any of the following form 0/0 , ∞/∞ then L'HOSPITAL'S RULE applied.

$\sf \implies \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}$

Note :

L'HOSPITAL'S RULE can be repeated required number of times in same questions.

Answer 2

FORMULA USED :-

$\longmapsto \boxed{ \green{ \bf \: \sf \lim_{x \to 0} \dfrac{sinx}{x} = 1}}$

$\longmapsto \boxed{ \green{ \bf \: sin(x + y) - sin(x - y) = 2 \: cosx \: siny}}$

CALCULATION :- .

$\tt \implies \lim_{x \to 0} \dfrac{sin(2 + x) - sin(2 - x)}{x}$

If we substitute the value of x = 0, we get indeterminant form.

So,

$\tt \implies \lim_{x \to 0} \dfrac{sin(2 + x) - sin(2 - x)}{x}$

$\tt \: = \: \lim_{x \to 0} \dfrac{2 \: cos2 \: sinx}{x}$

$\tt \: = \: 2 \: cos2 \: \lim_{x \to 0} \dfrac{sinx}{x}$

$\tt \: = \: 2 \: cos2 \: \times \: 1$

$\tt \: = \: 2 \: cos2$

$\boxed{ \green{ \bf \:Hence, \lim_{x \to 0} \dfrac{sin(2 + x) - sin(2 - x)}{x} = 2 \: cos2}}$

Additional Information :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\pink{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\purple{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} \: - \: 1}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\red{{\tt \:\lim_{x\to 0} \: \dfrac{ log(1 + x) }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\orange{{\tt \:\lim_{x\to 0} \: \dfrac{ {x}^{n} - {a}^{n} }{x \: - \: a} \: = \: {na}^{ n- 1} }}}}}} \\ \end{gathered}$