Lim_x→ 05^x 1/√(4+x) 2 =lim_x→ 0(5^x 1) ( √(4+x)+2 )/ (√(4+x) 2 ) (√(4+x)+2 ) =lim_x→ 0(5^x 1) ( √(4+x)+2 )/(4+x ) 4 =lim_x→ 0(5^x 1) ( √(4+x)+2 )/x =lim ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 05x-1...

Question

${lim}_{x \to 0} \frac{5^{x} - 1}{\sqrt{4 + x} - 2}$

Open in App

Solution

${lim}_{x \to 0} \frac{5^{x} - 1}{\sqrt{4 + x} - 2}$

$= {lim}_{x \to 0} \frac{(5^{x} - 1) (\sqrt{4 + x} + 2)}{(\sqrt{4 + x} - 2) (\sqrt{4 + x} + 2)}$

$= {lim}_{x \to 0} \frac{(5^{x} - 1) (\sqrt{4 + x} + 2)}{(4 + x) - 4}$

$= {lim}_{x \to 0} \frac{(5^{x} - 1) (\sqrt{4 + x} + 2)}{x}$

$= {lim}_{x \to 0} (\frac{5^{x} - 1}{x}) {lim}_{x \to 0} (\sqrt{4 + x} + 2)$

${∵ {lim}_{x \to 0} \frac{(a^{x} - 1)}{x} = l o g a}$

$= l o g 5 \times (\sqrt{4 + 0} + 2)$

$= 4 l o g 5$

Suggest Corrections

Similar questions

\lim_{x \to 0} \frac{5^{x} - 1}{\sqrt{4 + x} - 2}

Q. What is

lim x \to 0 \frac{5^{x} - 1}{x}

equal to?

${lim}_{x \to 0} \frac{s e c 5 x - s e c 3 x}{s e c 3 x - s e c x}$

${lim}_{x \to 0} \frac{5^{x} + 3^{x} + 2^{x} - 3}{x}$

\lim_{x \to 0} \frac{5 x + 4 \sin 3 x}{4 \sin 2 x + 7 x}

Join BYJU'S Learning Program

limx→05x−1√4+x−2

limx→05x−1√4+x−2 =limx→0(5x−1)(√4+x+2)(√4+x−2)(√4+x+2) =limx→0(5x−1)(√4+x+2)(4+x)−4 =limx→0(5x−1)(√4+x+2)x =limx→0(5x−1x)limx→0(√4+x+2) {∵limx→0(ax−1)x=log a} =log 5×(√4+0+2) =4 log5

${lim}_{x \to 0} \frac{5^{x} - 1}{\sqrt{4 + x} - 2}$