Question

1 Evaluate
lim sin2x-tan3x/x
x-0​

Answer 1

GIVEN :–

• A limit –

$\\ \implies \bf \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) - \tan(3x)}{x}\bigg\}$

TO FIND :–

• Value of limit = ?

SOLUTION :–

• Let –

$\\ \implies \bf P = \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) - \tan(3x)}{x}\bigg\}$

• Put the limit –

$\\ \implies \bf P =\dfrac{ \sin(2 \times 0) - \tan(3 \times 0)}{0}$

$\\ \implies \bf P =\dfrac{ \sin(0) - \tan(0)}{0}$

$\\ \implies \bf P =\dfrac{0-0}{0}$

$\\ \implies \bf P =\dfrac{0}{0}$

• It's undefined form. Now Let's solve –

$\\ \implies \bf P = \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x)}{x}\bigg\} -\lim_{x \to0} \bigg\{ \dfrac{ \tan(3x)}{x}\bigg\}$

• We can write this as –

$\\ \implies \bf P =\lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) \times2}{(2x)}\bigg\} -\lim_{x \to0} \bigg\{ \dfrac{ \tan(3x) \times 3}{3x}\bigg\}$

$\\ \implies \bf P =2\lim_{x \to0} \bigg\{ \dfrac{ \sin(2x)}{(2x)}\bigg\} -3\lim_{x \to0} \bigg\{ \dfrac{ \tan(3x)}{3x}\bigg\}$

• We also know that –

$\\ \implies \large\bf {\red{ \lim_{x \to0} \bigg\{ \dfrac{ \sin x}{x}\bigg\} = 1 \: \: , \: \: \lim_{x \to0} \bigg\{ \dfrac{ \tan x}{x}\bigg\} = 1}}$

• So that –

$\\ \implies \bf P =2(1) -3(1)$

$\\ \implies \bf P =2-3$

$\\ \implies \large{ \boxed{\bf P =-1}}$

• Hence –

$\\ \implies \large \pink{ \boxed{\bf \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) - \tan(3x)}{x}\bigg\} =-1}}$

Answer 2

Step-by-step explanation:

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$\text{\huge\underline{\red{Question:-}}}$

:$\Longrightarrow$ Evaluate

lim sin2x-tan3x/x

x-0.

$\text{\huge\underline{\orange{To\ find:-}}}$

:$\Longrightarrow$ ● We have to find the value of limit = ??

$\text{\huge\underline{\green{Given:-}}}$

:$\Longrightarrow$ ● Limit -

:$\Longrightarrow$ $\displaystyle \lim_{x\ to 0}\ \dfrac{sin(2x)-tan(3x)}{x}$

$\text{\huge\underline{\purple{Solution:-}}}$

:$\red{\bigstar}$ $\text{\large\underline{\bf{Firstly:-}}}$

:$\Longrightarrow$ P = $\displaystyle \lim_{x\ to 0}\ \dfrac{sin(2x)-tan(3x)}{x}$

:$\red{\bigstar}$ $\text{\large\underline{\bf{By\ putting\ the\ limit:-}}}$

:$\Longrightarrow$ P = $\mathrm{\dfrac{sin(2×0)-tan(3×0)}{0}}$

:$\Longrightarrow$ P = $\mathrm{\dfrac{sin(0)-tan(0)}{0}}$

:$\Longrightarrow$ P = $\mathrm{\dfrac{0-0}{0}}$

:$\Longrightarrow$ P = $\mathrm{\dfrac{0}{0}}$

:$\Longrightarrow$ ● So, it's undefined form.

:$\red{\bigstar}$ $\text{\Large\underline{\bf{So\ let's\ solve:-}}}$

:$\Longrightarrow$ P = $\displaystyle \lim_{x\ to 0}\ \dfrac{sin(2x)}{x}\ -\ \displaystyle \lim_{x\ to 0}\ \dfrac{tan(3x)}{x}$

:$\red{\bigstar}$ $\text{\Large\underline{\bf{We\ can\ also\ write\ it\ as:-}}}$

:$\Longrightarrow$ P = $\displaystyle \lim_{x\ to 0}\ \dfrac{sin(2x)×2}{(2x)}- \displaystyle \lim_{x\ to 0}\ \dfrac{tan(3x)×3}{3x}$

:$\Longrightarrow$ P = 2$\displaystyle \lim_{x\ to 0}\ \dfrac{sin(2x)}{(2x)}\ -3\ \displaystyle \lim_{x\ to 0}\ \dfrac{tan(3x)}{3x}$

:$\Longrightarrow$ ● We know that:-

:$\Longrightarrow$ $\displaystyle \lim_{x\ to 0}\ \dfrac{sin x}{x}\ =\ 1,\ \displaystyle \lim_{x\ to 0}\ \dfrac{tan x}{x}\ =\ 1$

:$\Longrightarrow$ ● So, that :-

:$\Longrightarrow$ P = 2(1) - 3(1)

:$\Longrightarrow$ P = 2 - 3

:$\Longrightarrow$ $\mathrm{P\ =\ -1}$

$\text{\Large\underline{\pink{Hence:-}}}$

:$\Longrightarrow$ $\displaystyle \lim_{x\ to 0}\ \dfrac{sin(2x)-tan(3x)}{x}\ =\ -1$

$\text{\huge\underline{\bf{Hence\ Verified}}}$

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