$lim x \to 0 \frac{a^{x} - 1}{b^{x} - 1}$ =

{log}_{b} a

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{log}_{a} b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{log}_{e} a b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{log}_{e} (\frac{a}{b})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A ${log}_{b} a$
$lim x \to 0 \frac{\frac{(a^{x} - 1)}{x}}{\frac{b^{x} - 1}{x}} = \frac{l o g a}{l o g b} = l o g_{b} a$

Suggest Corrections

Similar questions

2^{(\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}) . \sqrt{l o g_{a} b}}

= {\begin{matrix} 2, b \geq a \geq 1 2^{l o g_{a} b}, 1 < b < a \end{matrix}

[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0

log (\frac{a + b}{3}) = \frac{1}{2} (log a + log b)

l i m_{x \to 0} \frac{a^{x} - b^{x}}{e^{x} - 1}

A = {1, 2, 3, 4, 5, 6}

a, b, c ϵ A

lim x \to 0 {(\frac{a^{x} + b^{x} + c^{x}}{3})}^{\frac{3}{x}} = 6

\frac{1}{36}

lim x \to 0 {(\frac{a^{x} + b^{x} + c^{x}}{3})}^{\frac{3}{x}} = a b c

a > 0, b > 0, c > 0

Join BYJU'S Learning Program