Use the formula lim (x→0) (aˣ - 1)/x =...

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1982Easy

Use the formula $lim_{x \to 0} \frac{a ^{x} - 1}{x} = ln a$ to find $lim_{x \to 0} \frac{2 ^{x} - 1}{( 1 + x ) ^{1/2} - 1}$

Answer:

2 ln 2

Visualized Solution (Hindi)

Identifying the Limit Form

Evaluate: $lim_{x \to 0} \frac{2 ^{x} - 1}{( 1 + x ) ^{\frac{1}{2}} - 1}$
Check for indeterminate form by substituting $x = 0$ :
Numerator: $2^{0} - 1 = 1 - 1 = 0$
Denominator: $(1 + 0)^{\frac{1}{2}} - 1 = 1 - 1 = 0$
The expression is in the $\frac{0}{0}$ form.

The Standard Formula

Given formula: $lim_{x \to 0} \frac{a ^{x} - 1}{x} = ln a$
In our case, $a = 2$ .
We need to introduce $x$ in the denominator of the numerator to use this formula.

Manipulating the Expression

Divide numerator and denominator by $x$ :
$lim_{x \to 0} \frac{\frac{2 ^{x} - 1}{x}}{\frac{( 1 + x ) ^{\frac{1}{2}} - 1}{x}}$
This can be written as: $lim_{x \to 0} (\frac{2 ^{x} - 1}{x}) \cdot lim_{x \to 0} (\frac{x}{1 + x - 1})$

Evaluating the First Limit

Using the formula $lim_{x \to 0} \frac{a ^{x} - 1}{x} = ln a$ with $a = 2$ :
$lim_{x \to 0} \frac{2 ^{x} - 1}{x} = ln 2$

Handling the Denominator

Consider the second part: $lim_{x \to 0} \frac{x}{1 + x - 1}$
Rationalize the denominator by multiplying with $(1 + x + 1)$ :
$\frac{x ( 1 + x + 1 )}{( 1 + x - 1 ) ( 1 + x + 1 )}$

Simplifying the Fraction

Denominator becomes: $(1 + x)^{2} - (1)^{2} = 1 + x - 1 = x$
The expression simplifies to: $\frac{x ( 1 + x + 1 )}{x}$
Cancel $x$ from numerator and denominator: $1 + x + 1$

Evaluating the Second Limit

Apply the limit $x \to 0$ :
$lim_{x \to 0} (1 + x + 1) = 1 + 0 + 1$
$= 1 + 1 = 2$

The Final Result

Combine the results of both parts:
Total Limit $= ln 2 \cdot 2$
Final Answer: $2 ln 2$

00:00 / 00:00

Conceptually Similar Problems

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$lim_{x \to 0} \frac{x t a n 2 x - 2 x t a n x}{( 1 - c o s 2 x ) ^{2}}$ is

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Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

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Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

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MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2002Easy

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If $f (1) = 1, f^{'} (1) = 2$ , then $lim_{x \to 1} \frac{f ( x ) - 1}{x - 1}$ is

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(B)

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MathematicsEvaluation of Limits & L'Hopital's RuleJEE Advanced 1991Moderate

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The value of $lim_{x \to 0} \frac{\frac{1}{2} ( 1 - c o s 2 x )}{x}$

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$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

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Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Use the formula limx→0​xax−1​=lna to find limx→0​(1+x)1/2−12x−1​

Visualized Solution (Hindi)

Identifying the Limit Form

The Standard Formula

Manipulating the Expression

Evaluating the First Limit

Handling the Denominator

Simplifying the Fraction

Evaluating the Second Limit

The Final Result

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