Question

Evaluate.
Lim x tends to 0
Sin 2 x + 3x/2x + tan 3x​

Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 0} \dfrac{sin(2x) + 3x}{2x + tan(3x)}$

As we know that,

Put the value of x = 0 in equation and check their form.

$\sf \implies \lim_{x \to 0} \dfrac{sin(2(0)) + 3(0)}{2(0) + tan(3(0))}$

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

As we can see that it is the form of 0/0.

So, simply apply L HOSPITAL'S RULE in this equation, we get.

$\sf \implies \lim_{x \to 0} \dfrac{\dfrac{d}{dx} (sin(2x) + 3x)}{\dfrac{d}{dx} (2x + tan(3x))}$

$\sf \implies \lim_{x \to 0} \dfrac{2cos(2x) + 3}{2 + 3sec^{2}(3x) }$

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

$\sf \implies \lim_{x \to 0} \dfrac{2cos(2(0) + 3)}{2 + 3sec^{2} 3(0)}$

$\sf \implies \lim_{x \to 0} \dfrac{2sin(0) + 3}{2 + 3sec^{2} (0)}$

$\sf \implies \lim_{x \to 0} \dfrac{2 + 3}{2 + 3}$

$\sf \implies \lim_{x \to 0} \dfrac{5}{5} = 1$

                                                                                                                     

MORE INFORMATION.

Methods of evaluation of limits.

(1) = when $\lim_{x \to \infty}$

In this case expression should be expressed as a function 1/x and then after removing indeterminant form, (if it is there) replace 1/x by 0.

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

◉GIVEN:-

$lim_{x \to0} \: \: \frac{sin \: 2 x + 3x}{2x + tan 3x}$

◉LET'S PROVE:-

$lim_{x \to0} \: \frac{ \frac{2x(sin2x)}{2x + 3x} }{ \frac{2x + 3x(tan3x)}{3x} }$

$\bf{ \frac{lim2x_{x \to0} \: lim_{x \to0} \: [ \frac{sin2x}{2x}] \: +lim3x_{x \to0 }}{lim2x_{x \to0 \:lim3x_{x \to0 \: lim_{x \to0}}} \: [ \frac{tan \: 3x}{3x}] }}$

NOW AS,

$lim_{x \to0}[ \frac{tan \: 3x}{3x} ] \: and \: \: lim_{x \to0} \: \: [ \frac{sin \: 2x}{2x} ] \: both \: will \: \: be \: 1.$

$\leadsto \: \bf{ \frac{lim2x_{x \to0} \: lim_{x \to0} \: [ \frac{sin2x}{2x}] \: +lim3x_{x \to0 }}{lim2x_{x \to0 \:lim3x_{x \to0 \: lim_{x \to0}}} \: [ \frac{tan \: 3x}{3x}] }}$

$\leadsto \: \frac{lim2x.1_{x \to 0} \: + lim3x _{x \to0}}{lim2x_ {x \to0} + lim3x.1_ {x \to0}}$

$\leadsto \: lim_{x \to0} \: \frac{2x + 3x}{2x + 3x}$

$\leadsto \: lim1_{x \to0} \: = 1$

$: \longrightarrow \:{ \boxed{ \purple{ \tt{ lim_{x \to0} \: \: \frac{sin \: 2 x + 3x}{2x + tan 3x} = 1}}}}$

HOPE IT HELPS