Question

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Answer 1

We need to recall,

(i) $\displaystyle\lim_{x\to0}\dfrac{e^x-1}{x}=1$

(ii) $\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1$

(1). Given,

$\displaystyle\longrightarrow L_1=\lim_{x\to0}\dfrac{5^x-1}{\sin x}$

Since $a^b=e^{b\log a},$

$\displaystyle\longrightarrow L_1=\lim_{x\to0}\dfrac{e^{x\log5}-1}{\sin x}$

Divide numerator and denominator by $x.$

$\displaystyle\longrightarrow L_1 =\lim_{x\to0}\dfrac{\left(\dfrac{e^{x\log5}-1}{x}\right)}{\left( \dfrac{\sin x}{x}\right)}$

$\longrightarrow L_1=\dfrac{\displaystyle \lim_{x\to0}\dfrac{e^{x\log5}-1}{x}}{\displaystyle\lim_{x\to0}\dfrac{\sin x}{x} }$

$\longrightarrow L_1=\dfrac{\log5\displaystyle \lim_{(x\log5) \to0}\dfrac{e^{x\log5}-1}{x\log5}}{\displaystyle\lim_{x\to0}\dfrac{\sin x}{x} }$

By (i) and (ii),

$\longrightarrow L_1=\dfrac{\log5\cdot1}{1}$

$\longrightarrow\underline{\underline{L_1=\log5}}$

(2). Given,

$\longrightarrow L_2=\displaystyle\lim_{x\to0}\dfrac{5^x-4^x}{x}$

$\longrightarrow L_2=\displaystyle\lim_{x\to0}\dfrac{e^{x\log5}-e^{x\log4}}{x}$

$\longrightarrow L_2=\displaystyle\lim_{x\to0}\dfrac{e^{x\log5}-1}{x}-\displaystyle\lim_{x\to0}\dfrac{e^{x\log4}-1}{x}$

$\longrightarrow L_2=\log5\displaystyle\lim_{(x\log5)\to0}\dfrac{e^{x\log5}-1}{x\log5}-\log4\displaystyle\lim_{(x\log4)\to0}\dfrac{e^{x\log4}-1}{x\log4}$

By (i),

$\longrightarrow L_2=\log5\cdot1-\log4\cdot1$

$\longrightarrow\underline{\underline{L_2=\log\left(\dfrac{5}{4}\right)}}$

(3). Perform the same logic of (2) and we get,

$\longrightarrow\underline{\underline{L_3=\log\left(\dfrac{5}{3}\right)}}$

(4). Given,

$\longrightarrow L_4=\displaystyle\lim_{x\to0}\dfrac{4^x-1}{3^x-1}$

$\longrightarrow L_4=\displaystyle\lim_{x\to0}\dfrac{e^{x\log4}-1}{e^{x\log3}-1}$

Divide numerator and denominator by $x.$

$\longrightarrow L_4=\displaystyle\lim_{x\to0}\dfrac{\left(\dfrac{e^{x\log4}-1}{x}\right)}{\left(\dfrac{e^{x\log3}-1}{x}\right)}$

$\longrightarrow L_4=\dfrac{\displaystyle\lim_{x\to0}\dfrac{e^{x\log4}-1}{x}}{\displaystyle\lim_{x\to0}\dfrac{e^{x\log3}-1}{x}}$

$\longrightarrow L_4=\dfrac{\log4\displaystyle\lim_{(x\log4)\to0}\dfrac{e^{x\log4}-1}{x\log4}}{\log3\displaystyle\lim_{(x\log3)\to0}\dfrac{e^{x\log3}-1}{x\log3}}$

By (i),

$\longrightarrow L_4=\dfrac{\log4\cdot1}{\log3\cdot1}$

$\longrightarrow\underline{\underline{L_4=\log_34}}$