Question

Solve the given limits :-
$\rm \lim \limits_{x \to1} \left \{ \dfrac{(x + 1)^{4}-{2}^{4} }{(2x + 1)^{5} - {3}^{5} } \right \}$

Don't use binomial expansion, if possible try to solve using L'Hopital rule.​

Answer 1

Answer:

16/405 will be the correct answer

Step-by-step explanation:

see steps in attachment

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\rm \lim \limits_{x \to1} \left \{ \dfrac{(x + 1)^{4}-{2}^{4} }{(2x + 1)^{5} - {3}^{5} } \right \}$

If we put directly x = 1, we get

$\rm \:  =  \: \left \{ \dfrac{(1 + 1)^{4}-{2}^{4} }{(2 + 1)^{5} - {3}^{5} } \right \}$

$\rm \:  =  \: \left \{ \dfrac{(2)^{4}-{2}^{4} }{(3)^{5} - {3}^{5} } \right \}$

$\rm \:  =  \: \dfrac{0}{0} \: which \: is \: indeterminant$

So, to evaluate the limit,

$\rm \:  =  \: \rm \lim \limits_{x \to1} \left \{ \dfrac{(x + 1)^{4}-{2}^{4} }{(2x + 1)^{5} - {3}^{5} } \right \}$

By using L - Hospital Rule,

$\rm \:  =  \: \rm \lim \limits_{x \to1} \left \{ \dfrac{\dfrac{d}{dx} (x + 1)^{4}-\dfrac{d}{dx}{2}^{4} }{\dfrac{d}{dx}(2x + 1)^{5} - \dfrac{d}{dx}{3}^{5} } \right \}$

We know that,

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx}k = 0}}}$

and

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}}$

So, using these Identities, we get

$\rm \:  =  \: \rm \lim \limits_{x \to1} \left \{ \dfrac{4(x + 1)^{3}\dfrac{d}{dx}(x + 1)}{5(2x + 1)^{4}\dfrac{d}{dx}(2x + 1)} \right \}$

$\rm \:  =  \: \rm \lim \limits_{x \to1} \left \{ \dfrac{4(x + 1)^{3}(1 + 0)}{5(2x + 1)^{4}(2 + 0)} \right \}$

$\rm \:  =  \: \rm \lim \limits_{x \to1} \left \{ \dfrac{4(x + 1)^{3}}{10(2x + 1)^{4}} \right \}$

$\rm \:  =  \:\dfrac{2(1+ 1)^{3}}{5(2 + 1)^{4}}$

$\rm \:  =  \:\dfrac{2(2)^{3}}{5(3)^{4}}$

$\rm \:  =  \: \dfrac{2 \times 8}{5 \times 81}$

$\rm \:  =  \: \dfrac{16}{405}$

Hence,

$\bf\implies \:\red{ \boxed{ \sf{ \:\rm \lim \limits_{x \to1} \left \{ \dfrac{(x + 1)^{4}-{2}^{4} }{(2x + 1)^{5} - {3}^{5} } \right \} = \frac{16}{405}}}}$

Additional Information :-

$\blue{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$