Evaluate the given: [tex]\lim_{x \to 0}\:\: \dfrac{10^x - 2^x - 5^x + 1}{x.Tanx}[/tex] Needed Step by Step Solution. Class XI:- Limits and Derivatives. Created by Steph0303. Math

tex] Needed Step by Step Solution. Class XI:- Limits and Derivatives...

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Answer:

Considering the function, we simplify to get,

$\implies \dfrac{ 5^x.2^x - 2^x - 5^x + 1}{x.Tanx}\\\\\implies \dfrac{ 2^x\:[\: 5^x - 1\:]\: -1 \: [ \: 5^x - 1\:]}{x.Tanx}\\\\\implies \dfrac{ [ 2^x - 1 ] [ 5^x - 1 ]}{x.Tanx}\\\\Multiplying\:and\:dividing\:by\:x\:we\:get,\\\\\implies \dfrac{ [ 2^x - 1 ] [ 5^x - 1 ]}{x.Tanx} \times \dfrac{x}{x}\\\\\implies \dfrac{ \:x[ 2^x - 1 ] [ 5^x - 1 ]}{x^2} \times \dfrac{x}{Tanx}$

$Now\:Applying\:limits\:we\:get,\\\\\implies \lim_{x \to 0}\:\: \dfrac{ \:x[\ 2^x - 1 ] [ 5^x - 1 ]}{x^2} .\: \lim_{x \to 0}\:\: \dfrac{x}{Tanx}\\\\\\We\:know\:that\: \dfrac{x}{Tanx}\: when\: x \to 0\:is\:1\\\\\implies \lim_{x \to 0}\:\: \dfrac{\:x [ 2^x - 1 ] [ 5^x - 1 ]}{x^2} .\: 1$

x and x² cancel each other and we get x in the denominator. Hence we get,

$\implies \lim_{x \to 0}\:\: \dfrac{ [ 2^x - 1 ] [ 5^x - 1 ]}{x}\\\\We\:know\:that\:\: \dfrac{a^x - 1}{x} = Ln\:a\\\\\implies \lim_{x \to 0}\:\: \dfrac{ [ 2^x - 1 ]}{x} = Ln\:2\\\\\implies \lim_{x \to 0}\:\: \dfrac{[ 5^x - 1 ]}{x} = Ln\:5\\\\\implies \lim_{x \to 0}\:\: \dfrac{ [ 2^x - 1 ] [ 5^x - 1 ]}{x} = Ln\:2 \:\:.\:\:Ln\:5$

This is the needed answer.

Hope I'm correct.

Steph0303 Thanks a lot :)

Pikaachu

Answer:

$ln(2) . ln(5)$

Step by Step Explanation :

Start of by Factorising :

$\lim_{x \to0} \: \frac{ {10}^{x} - {5}^{x} - {2}^{x} + 1}{x \tan x }$

$= \lim_{x \to0} \: \frac{ ({5}^{x} - 1)( {2}^{x} - 1)}{ {x}^{2} } \times \frac{x}{ \tan x }$

$= \lim_{x \to0} \: \frac{ ({5}^{x} - 1)}{ x} \times \frac{ ({2}^{x} - 1)}{ x} \times \frac{x}{ \tan x }$

Using :

$\lim_{x \to0} \: \frac{ {a}^{x} - 1}{x} = ln(a) ; \: \: \\ \\ \lim_{x \to0} \: \frac{ x}{ \tan x} = 1$

So, we have our expression converted as :

$= \lim_{x \to0} \: \frac{ ({5}^{x} - 1)}{ x} \times \frac{ ({2}^{x} - 1)}{ x} \times \frac{x}{ \tan x }$

$= ln(2) . ln(5)$

∆ L'hopital would give a worse look, so there's no better soln to this one.

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