Question

lim x tends to 0( (1+x)^5 -1) /(3x+5x^2)​

Answer 1

Step-by-step explanation:

We have,

$\lim _{x \rarr0} \frac{ {(1 + x)}^{5} - 1}{3x + 5 {x}^{2} }$

This is the form of 0/0, so,

using l'hospital rule,

$\lim _{x \rarr0} \frac{ \frac{d}{dx}((1 + x)^{5} - 1) }{ \frac{d}{dx}(3x + 5 {x}^{2} )}$

$= \lim _{x \rarr0} \frac{5(1 + x) ^{4} }{3 + 10x}$

$= \frac{5(1 + 0) ^{4} }{3 + 10(0)}$

$= \frac{5 \times 1}{3 + 0}$

$= \frac{5}{3}$

Answer 2

$\rm \longrightarrow \: \displaystyle \lim_ {x \to \: 0} \rm{ \dfrac{ {(1 + x)}^{5} - 1 }{3x + 5 {x}^{2} } }$

If we put x=0 in the above expression we get 0/0 form. Since 0/0 is indeterminate form , So we will apply L'Hospital rule to solve this.

While applying L'Hospital rule , we have to differentiate numerator and denominator seperately.

$\rm \longrightarrow \: \displaystyle \lim_ {x \to \: 0} \rm{ \dfrac{d \bigg(\dfrac{ {(1 + x)}^{5} - 1}{dx} \bigg)}{ \dfrac{d(3x + 5 {x}^{2})}{dx} } }$

$\rm \longrightarrow \: \displaystyle \lim_ {x \to \: 0} \rm{ \dfrac{ {5(1 + x)}^{4} - 0 }{3+ 5 {x}^{} } }$

$\rm \longrightarrow \: \displaystyle \lim_ {x \to \: 0} \rm{ \dfrac{ {5(1 + x)}^{4} }{3+ 5 {x}^{} } }$

Now put x= 0.

$\rm \longrightarrow \: \displaystyle \rm{ \dfrac{ {5(1 + 0)}^{4} }{3+ (5 \times 0)} }$

$\rm \longrightarrow \: \displaystyle \rm{ \dfrac{ {5(1)}^{4} }{3+ 0} }$

$\rm \longrightarrow \: \displaystyle \rm{ \dfrac{ {5} }{3} }$

Answer : 5/3