Question

plz solve part 4 of this question...​

Answer 1

Answer:

$\large\bold\red{ln(a) ln(b) }$

Step-by-step explanation:

Given,

$(iv)\lim_{x \to 0 } \frac{ {(ab)}^{x} - {a}^{x} - {b}^{x} + 1 }{ {x}^{2} }$

Further simplifying,

we get,

$= \lim_{x \to 0 } \frac{ {a}^{x} {b}^{x} - {a}^{x} - {b}^{x} + 1 }{ {x}^{2} }$

Now,

doing factorisation,

we get,

$= \lim_{x \to 0 } \frac{ {a}^{x} ( {b}^{x} - 1) - 1( {b}^{x} - 1)}{ {x}^{2} } \\ \\ = \lim_{x \to 0 } \frac{( {a}^{x} - 1)( {b}^{x} - 1)}{ {x}^{2} } \\ \\ = \lim_{x \to 0 } \frac{ {a}^{x} - 1}{x} \times \lim_{x \to 0 } \frac{ {b}^{x} - 1}{x}$

But,

we know that,

$\lim_{h \to 0 } \frac{ {m}^{h} - 1}{h} = ln(m)$

Therefore,

we get,

$= \bold{ln(a) ln(b) }$