Question

Evaluate this

lim x → 0 [(1+x)⁶-1]/[(1+x)²-1]​

Answer 1

Given

$\to \displaystyle \lim_{ \tt \: x \to0} \tt \dfrac{(1 + x)^{6} - 1 }{(1 + x) {}^{2} - 1}$

Now Check the form so put x = 0 on Given equation

$\tt \to \: \dfrac{(1 + x) ^{6} - 1 }{(1 + x) {}^{2} - 1}$

$\tt \to \: \dfrac{(1 + 0) ^{6} - 1 }{(1 + 0) {}^{2} - 1}$

$\tt \to \: \dfrac{1 - 1 }{1 - 1} = \dfrac{0}{0}$

So it's 0/0 form we use L'hospital rule

$\to \displaystyle \lim_{ \tt \: x \to0} \tt \dfrac{ \dfrac{d \{(1 + x)^{6} - 1 \}}{dx} }{ \dfrac{d \{(1 + x) {}^{2} - 1 \}}{dx}}$

Now Using Chain Rule we get

$\to \displaystyle \lim_{ \tt \: x \to0} \tt \frac{6(1 + x) ^{6- 1} \times \dfrac{d(1 + x)}{dx} }{2(1 + x) ^{2 - 1} \times \dfrac{d(1 + x)}{dx} }$

$\to \displaystyle \lim_{ \tt \: x \to0} \tt \frac{6(1 + x) ^{5} \times 1 }{2(1 + x) \times 1}$

$\tt \to \: \dfrac{6(1 + 0) {}^{5} }{2(1 + 0)}$

$\tt \to \: \dfrac{6(1) {}^{5} }{2(1)} = \dfrac{6}{2} = 3$

Answer is 3