Let $p = \lim_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$ then $\log p$ is equal to Options: (A) 1/2 (B) 1/4 (C) 2 (D) 1

### Analyzing the Setup We are given the limit $p = \lim_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$ and our mission is to find $\log p$. First, let's perform the most important step in any limit problem: direct substitution. As $x \to 0^+$, $\sqrt{x} \to 0$, and $\tan(0) = 0$. Thus, the base becomes $1 + 0 = 1$. Meanwhile, the exponent $\frac{1}{2x}$ approaches $\infty$. We have arrived at the **$1^{\infty}$ indeterminate form**. ### The Master Equation This is a classic scenario in calculus. Whenever you encounter this form, we use the standard identity: $$\lim_{x \to a} (1 + f(x))^{g(x)} = e^{\lim_{x \to a} f(x)g(x)}$$ Here, we identify $f(x) = \tan^2 \sqrt{x}$ and $g(x) = \frac{1}{2x}$. Substituting these into the identity, we get: $$p = e^{\lim_{x \to 0^+} \frac{\tan^2 \sqrt{x}}{2x}}$$ ### Final Calculation The problem asks for $\log p$. Taking the natural logarithm of both sides, we obtain: $$\log p = \lim_{x \to 0^+} \frac{\tan^2 \sqrt{x}}{2x}$$ To solve this, we recall the standard limit $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$. We can rewrite our expression as: $$\log p = \frac{1}{2} \lim_{x \to 0^+} \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right)^2$$ As $x \to 0^+$, $\sqrt{x} \to 0$, so the limit inside the bracket becomes $1$. Thus, the final result is: $$\log p = \frac{1}{2} (1)^2 = \mathbf{\frac{1}{2}}$$

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

Q: Wait, why is the base $1$ and the exponent $\infty$?

Great observation! As $x \to 0^+$, $\sqrt{x}$ also approaches $0$. Since $\tan(0) = 0$, the base $(1 + \tan^2 \sqrt{x})$ becomes $(1 + 0) = 1$. Simultaneously, the exponent $\frac{1}{2x}$ becomes $\frac{1}{2(0^+)} = \infty$. This confirms we are dealing with the $1^{\infty}$ indeterminate form.

Q: Why do we take $\log$ at the end?

The question specifically asks for $\log p$. By taking the natural logarithm of both sides, we use the property $\log(e^y) = y$. This effectively 'brings down' the exponent, allowing us to solve the limit of the expression $\frac{\tan^2 \sqrt{x}}{2x}$ directly.

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Visualized Solution

The Limit Expression

Given: $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$
We need to find the value of $lo g p$ .

Checking the Limit Form

Substitute $x \to 0^{+}$ into the base and exponent.
Base: $1 + tan^{2} 0 = 1 + 0 = 1$
Exponent: $\frac{1}{2 ( 0 ^{+} )} \to \infty$

The $1^{\infty}$ Indeterminate Form

The limit takes the form $1^{\infty}$ .
This is a standard indeterminate form in calculus.

Formula for $1^{\infty}$ Form

If $lim_{x \to a} f (x) = 0$ and $lim_{x \to a} g (x) = \infty$
Then $lim_{x \to a} (1 + f (x))^{g (x)} = e^{l i m_{x \to a} (f (x) \cdot g (x))}$

Applying the Formula

Identify $f (x)$ and $g (x)$ from our problem:
$f (x) = tan^{2} x$
$g (x) = \frac{1}{2 x}$

Setting up the Exponential Limit

Substitute $f (x)$ and $g (x)$ into the exponent:
$p = e^{l i m_{x \to 0^{+}} (t a n^{2} x \cdot \frac{1}{2 x})}$

Simplifying the Expression

Combine the terms in the exponent:
$p = e^{l i m_{x \to 0^{+}} \frac{t a n ^{2} x}{2 x}}$

Taking Logarithm on Both Sides

The question asks for $lo g p$ .
Take the natural logarithm ( $lo g_{e}$ or $ln$ ) on both sides:
$lo g p = lo g (e^{l i m_{x \to 0^{+}} \frac{t a n ^{2} x}{2 x}})$

Simplifying the Logarithm

Use the property $lo g (e^{y}) = y$ :
$lo g p = lim_{x \to 0^{+}} \frac{t a n ^{2} x}{2 x}$

Standard Trigonometric Limit

Recall the standard limit: $lim_{θ \to 0} \frac{t a n θ}{θ} = 1$
We need to manipulate our expression to match this form.

Rearranging the Expression

Rewrite $2 x$ as $2 (x)^{2}$ :
$lo g p = lim_{x \to 0^{+}} \frac{t a n ^{2} x}{2 ( x ) ^{2}}$
Pull out the constant $\frac{1}{2}$ :
$lo g p = \frac{1}{2} lim_{x \to 0^{+}} (\frac{t a n x}{x})^{2}$

Final Calculation

As $x \to 0^{+}$ , $x \to 0$ .
Apply the standard limit: $lim_{x \to 0} \frac{t a n x}{x} = 1$
$lo g p = \frac{1}{2} (1)^{2} = \frac{1}{2}$

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The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

We are given the limit

p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}

and our mission is to find

lo g p

First, let's perform the most important step in any limit problem: direct substitution. As

x \to 0^{+}

x \to 0

, and

tan (0) = 0

. Thus, the base becomes

1 + 0 = 1

Meanwhile, the exponent

\frac{1}{2 x}

approaches

\infty

. We have arrived at the $1^{\infty}$ indeterminate form.

The Master Equation

This is a classic scenario in calculus. Whenever you encounter this form, we use the standard identity:

x \to a lim (1 + f (x))^{g (x)} = e^{l i m_{x \to a} f (x) g (x)}

Here, we identify

f (x) = tan^{2} x

and

g (x) = \frac{1}{2 x}

. Substituting these into the identity, we get:

p = e^{l i m_{x \to 0^{+}} \frac{t a n ^{2} x}{2 x}}

Final Calculation

The problem asks for

lo g p

. Taking the natural logarithm of both sides, we obtain:

lo g p = x \to 0^{+} lim \frac{tan ^{2} x}{2 x}

To solve this, we recall the standard limit

lim_{θ \to 0} \frac{t a n θ}{θ} = 1

. We can rewrite our expression as:

lo g p = \frac{1}{2} x \to 0^{+} lim (\frac{tan x}{x})^{2}

x \to 0^{+}

x \to 0

, so the limit inside the bracket becomes

1

. Thus, the final result is:

lo g p = \frac{1}{2} (1)^{2} = \frac{1}{2}

Common Traps & FAQs

Wait, why is the base

1

and the exponent

\infty

Great observation! As

x \to 0^{+}

x

also approaches

0

. Since

tan (0) = 0

, the base

(1 + tan^{2} x)

becomes

(1 + 0) = 1

. Simultaneously, the exponent

\frac{1}{2 x}

becomes

\frac{1}{2 ( 0 ^{+} )} = \infty

. This confirms we are dealing with the

1^{\infty}

indeterminate form.

Why do we take

lo g

at the end?

The question specifically asks for

lo g p

. By taking the natural logarithm of both sides, we use the property

lo g (e^{y}) = y

. This effectively 'brings down' the exponent, allowing us to solve the limit of the expression

\frac{t a n ^{2} x}{2 x}

directly.

Can I use L'Hopital's rule instead of the standard limit?

Absolutely! L'Hopital's rule is a powerful tool. If you differentiate the numerator and denominator with respect to

x

, you will arrive at the same result. However, using the standard limit

lim_{θ \to 0} \frac{t a n θ}{θ} = 1

is often faster and demonstrates a deeper understanding of trigonometric limits.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

Select Answer:

Visualized Solution

The Limit Expression

Checking the Limit Form

The $1^{\infty}$ Indeterminate Form

Formula for $1^{\infty}$ Form

Applying the Formula

Setting up the Exponential Limit

Simplifying the Expression

Taking Logarithm on Both Sides

Simplifying the Logarithm

Standard Trigonometric Limit

Rearranging the Expression

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Master Equation

Final Calculation

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p=limx→0+​(1+tan2x​)2x1​ then logp is equal to

Select Answer:

Visualized Solution

The Limit Expression

Checking the Limit Form

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The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

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The Master Equation

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

The $1^{\infty}$ Indeterminate Form

Formula for $1^{\infty}$ Form

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