Question

Evaluate the limit
$\lim_{x \rightarrow 1}\frac{(x+1)^{4} - 2^4}{(2x+1)^{5} - 3^5}$

Answer 1

ANSWER:

To Evaluate:

$\:\:\bullet\:\:\lim_{x \to 1}\dfrac{(x+1)^4-2^4}{(2x+1)^5-3^5}$

Solution:

We need to evaluate that,

$\implies\lim_{x \to 1}\dfrac{(x+1)^4-2^4}{(2x+1)^5-3^5}$

First we will substitute value of x as 1. So,

$\implies\lim_{x \to 1}\dfrac{(x+1)^4-2^4}{(2x+1)^5-3^5}$

$\implies\dfrac{(1+1)^4-2^4}{(2(1)+1)^5-3^5}$

$\implies\dfrac{2^4-2^4}{3^5-3^5}$

$\implies\dfrac{0}{0}$

This is the 0/0 form, which is indeterminate one. So, we apply L'Hôpital Rule.

According to L'Hôpital Rule, if

$\implies\lim_{x \to a}\dfrac{f(x)}{g(x)}=\dfrac{0}{0}$

Or,

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

Then,

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

We had,

$\implies\lim_{x \to 1}\dfrac{(x+1)^4-2^4}{(2x+1)^5-3^5}=\dfrac{0}{0}$

So, here

• $a = 1$
• $f(x) = (x+1)^4 - 2^4$
• $g(x) = (2x+1)^5 - 3^5$

Hence,

$\implies\lim_{x \to 1}\dfrac{(x+1)^4-2^4}{(2x+1)^5-3^5}=\lim_{x \to 1}\dfrac{f(x)}{g(x)}$

Now, we'll calculate the derivatives of f(x) and g(x), i.e., f'(x) and g'(x).

Taking f(x) first,

$\implies f(x)=(x+1)^4-2^4$

$\implies y=(x+1)^4-16$

On dfferentiating,

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg[(x+1)^4-16\bigg]$

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg[(x+1)^4\bigg]+\dfrac{d}{dx}\bigg[-16\bigg]$

So,

$\implies \dfrac{dy}{dx}=4(x+1)^3\times\dfrac{d}{dx}\bigg[(x+1)\bigg]+0$

$\implies \dfrac{dy}{dx}=4(x+1)^3\times 1$

$\implies f'(x)=4(x+1)^3- - - - (1)$

Now, taking g(x),

$\implies g(x)=(2x+1)^5-3^5$

$\implies y=(2x+1)^5-243$

On dfferentiating,

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg[(2x+1)^5-243\bigg]$

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg[(2x+1)^5\bigg]+\dfrac{d}{dx}\bigg[-243\bigg]$

So,

$\implies \dfrac{dy}{dx}=5(2x+1)^4\times\dfrac{d}{dx}\bigg[(2x+1)\bigg]+0$

$\implies \dfrac{dy}{dx}=5(2x+1)^4\times 2$

$\implies g'(x)=10(2x+1)^4- - - - (2)$

So, we had,

$\implies\lim_{x \to 1}\dfrac{(x+1)^4-2^4}{(2x+1)^5-3^5}=\lim_{x \to 1}\dfrac{f(x)}{g(x)}$

$\implies\lim_{x \to 1}\dfrac{f(x)}{g(x)}$

Using, L'Hôpital Rule,

$\implies\lim_{x \to 1}\dfrac{f'(x)}{g'(x)}$

From (1) and (2),

$\implies\lim_{x \to 1}\dfrac{4\!\!\!/^{\:2}\,(x+1)^3}{10\!\!\!\!\!/_{\:\:5}\,(2x+1)^4}$

$\implies\lim_{x \to 1}\dfrac{2(x+1)^3}{5(2x+1)^4}$

Substituting the value of x as 1,

$\implies\dfrac{2(1+1)^3}{5(2(1)+1)^4}$

$\implies\dfrac{2(2)^3}{5(3)^4}$

$\implies\dfrac{2\times8}{5\times81}$

$\implies\bf\dfrac{16}{343}$

Therefore,

$\implies\bf\lim_{x \to 1}\dfrac{(x+1)^4-2^4}{(2x+1)^5-3^5}=\dfrac{16}{343}$