Evaluate $\lim_{x \to 0} \dfrac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x}$.

> [!TLDR] > This limit is a $\frac{0}{0}$ indeterminate form. By rationalising the numerator and using the identity $1 - \cos x = 2\sin^2(x/2)$, the limit simplifies to $\frac{1}{4\sqrt{2}}$. When you first look at $\lim_{x \to 0} \dfrac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x}$, the immediate instinct is to substitute $x = 0$. Doing that gives $\frac{\sqrt{2} - \sqrt{1+1}}{0} = \frac{0}{0}$. That’s an indeterminate form, so we need to manipulate the expression. The core idea: whenever you see a difference of square roots, rationalising the numerator is almost always the first step. It removes the square roots and often reveals a factor that cancels with the denominator. Here, the denominator is $\sin^2 x$, which near $x=0$ behaves like $x^2$, so we expect the final answer to be a finite number. Let’s work through it step by step. 1. **Rationalise the numerator.** Multiply numerator and denominator by the conjugate $\sqrt{2} + \sqrt{1 + \cos x}$: \[ \frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x} \cdot \frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}} = \frac{2 - (1 + \cos x)}{\sin^2 x \, (\sqrt{2} + \sqrt{1 + \cos x})} \] The numerator simplifies to $2 - 1 - \cos x = 1 - \cos x$. 2. **Rewrite the numerator using a half-angle identity.** The identity $1 - \cos x = 2\sin^2(x/2)$ is a standard tool for limits. So: \[ \frac{1 - \cos x}{\sin^2 x \, (\sqrt{2} + \sqrt{1 + \cos x})} = \frac{2\sin^2(x/2)}{\sin^2 x \, (\sqrt{2} + \sqrt{1 + \cos x})} \] 3. **Handle the denominator $\sin^2 x$.** For small $x$, $\sin x \sim x$, but we can be precise using the double-angle identity: $\sin x = 2\sin(x/2)\cos(x/2)$. Then $\sin^2 x = 4\sin^2(x/2)\cos^2(x/2)$. Substitute: \[ \frac{2\sin^2(x/2)}{4\sin^2(x/2)\cos^2(x/2) \, (\sqrt{2} + \sqrt{1 + \cos x})} = \frac{2}{4\cos^2(x/2) \, (\sqrt{2} + \sqrt{1 + \cos x})} \] Cancel $\sin^2(x/2)$ (valid for $x \neq 0$, and the limit as $x \to 0$ is unaffected). This simplifies to: \[ \fra

Evaluate limx → 0 dfrac√(2) - √(1 + cos x)sin² x. – Class 11 Mathematics

NCERT Exemplar · Q26

\lim_{x \to 0} \dfrac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x}

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Concept understanding — Standard Limits

A Toolkit of Named Limits

Some limits recur so often across problems that it is worth memorising their values outright, along with the one theorem that proves the trickiest of them: the Sandwich (Squeeze) Theorem.

The Sandwich Theorem

Theorem 9.5. If $g(x) \le f(x) \le h(x)$ for all $x$ near $x_0$ (except possibly at $x_0$ itself), and if

\lim_{x\to x_0} g(x) = \lim_{x\to x_0} h(x) = l,

then $\lim_{x\to x_0} f(x) = l$ too — $f$ is "squeezed" between two functions that agree in the limit, so it has no room to do anything else.

Illustration: to show $\displaystyle\lim_{x\to 0} x^2\sin\!\frac{1}{x^2} = 0$ , note that $\sin(\cdot)$ is always between $-1$ and $1$ , so $-x^2 \le x^2\sin\frac{1}{x^2} \le x^2$ . Since both $-x^2$ and $x^2$ tend to $0$ as $x\to 0$ , the Sandwich Theorem forces the middle expression to $0$ as well — even though $\lim_{x\to 0}\sin\frac{1}{x^2}$ on its own does not exist (it oscillates wildly), so the product rule alone could never have been applied directly.

Watch out

This is exactly why the Sandwich Theorem is indispensable rather than a curiosity: whenever one factor oscillates without a limit but is bounded, and the other factor is squeezed to zero, the ordinary product law (Concept 2) is not applicable — you need the sandwich.

The two flagship trigonometric limits

Result 9.1.

\text{(a)}\quad \lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1 \qquad\qquad \text{(b)}\quad \lim_{\theta\to 0}\frac{1-\cos\theta}{\theta} = 0

Part (a) is proved geometrically by sandwiching $\frac{\sin\theta}{\theta}$ between $\cos\theta$ and $1$ using the areas of a triangle, a sector, and a larger triangle built on the unit circle; since both bounding functions tend to $1$ as $\theta\to0$ , so must $\frac{\sin\theta}{\theta}$ . Part (b) follows algebraically from (a) by writing $1-\cos\theta = 2\sin^2\frac{\theta}{2}$ and splitting the quotient into a $\sin$ -over-argument piece (which uses part (a)) times a factor that vanishes.

A direct corollary worth keeping separate: $\displaystyle\lim_{x\to 0}\sin x = 0$ , obtained from the sandwich $-|x|\le \sin x \le |x|$ .

The full standard-limit toolkit (§9.2.10)

Alongside the trig pair above, these are worth having on instant recall — none require anything beyond algebra and substitution to use (their proofs, where given, lean on the exponential/log relationship or on the trig pair):

\lim_{x\to 0}\frac{e^x-1}{x}=1 \qquad\quad \lim_{x\to 0}\frac{a^x-1}{x}=\log a \ (a>0) \qquad\quad \lim_{x\to 0}\frac{\log(1+x)}{x}=1

\lim_{x\to 0}\frac{\sin^{-1}x}{x}=1 \qquad\quad \lim_{x\to 0}\frac{\tan^{-1}x}{x}=1

And the three equivalent forms of the number $e$ as a limit:

\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e \qquad\quad \lim_{x\to 0}(1+x)^{1/x} = e \qquad\quad \lim_{x\to\infty}\left(1+\frac{k}{x}\right)^x = e^k

Note

$e$ is a transcendental number — it never satisfies any polynomial equation with rational coefficients. That's part of why it shows up as a genuinely new limiting constant here rather than something expressible in simpler closed form.

The recognise-and-substitute pattern

Nearly every "hard-looking" limit in this section is really one of the above standard forms in disguise, reached via a clean substitution $y = (\text{some expression in } x)$ chosen so that $y \to 0$ (or $y\to\infty$ ) exactly when $x$ does. The book's worked examples all follow this shape:

Spot the shell. Identify which standard form the expression resembles — a $\left(1+\Box\right)^{1/\Box}$ shape signals $e$ ; a $\dfrac{\sin(\Box)}{\Box}$ shape signals Result 9.1(a); a $\dfrac{a^{\Box}-1}{\Box}$ shape signals Result 9.3.
Substitute $y$ for the " $\Box$ " so the expression matches the standard form exactly, tracking what $y \to$ as $x \to x_0$ .
Apply the standard limit to the $y$ -expression, then (if the exponent or coefficient outside doesn't vanish) combine using the power/product rules from Concept 2. …

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This limit is a $\frac{0}{0}$ indeterminate form. By rationalising the numerator and using the identity $1 - \cos x = 2\sin^2(x/2)$ , the limit simplifies to $\frac{1}{4\sqrt{2}}$ .

When you first look at $\lim_{x \to 0} \dfrac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x}$ , the immediate instinct is to substitute $x = 0$ . Doing that gives $\frac{\sqrt{2} - \sqrt{1+1}}{0} = \frac{0}{0}$ . That’s an indeterminate form, so we need to manipulate the expression.

The core idea: whenever you see a difference of square roots, rationalising the numerator is almost always the first step. It removes the square roots and often reveals a factor that cancels with the denominator. Here, the denominator is $\sin^2 x$ , which near $x=0$ behaves like $x^2$ , so we expect the final answer to be a finite number.

Let’s work through it step by step.

Rationalise the numerator. Multiply numerator and denominator by the conjugate $\sqrt{2} + \sqrt{1 + \cos x}$ :

\frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x} \cdot \frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}} = \frac{2 - (1 + \cos x)}{\sin^2 x \, (\sqrt{2} + \sqrt{1 + \cos x})}

The numerator simplifies to $2 - 1 - \cos x = 1 - \cos x$ .

Rewrite the numerator using a half-angle identity. The identity $1 - \cos x = 2\sin^2(x/2)$ is a standard tool for limits. So:

\frac{1 - \cos x}{\sin^2 x \, (\sqrt{2} + \sqrt{1 + \cos x})} = \frac{2\sin^2(x/2)}{\sin^2 x \, (\sqrt{2} + \sqrt{1 + \cos x})}

Handle the denominator $\sin^2 x$ . For small $x$ , $\sin x \sim x$ , but we can be precise using the double-angle identity: $\sin x = 2\sin(x/2)\cos(x/2)$ . Then $\sin^2 x = 4\sin^2(x/2)\cos^2(x/2)$ . Substitute:

\frac{2\sin^2(x/2)}{4\sin^2(x/2)\cos^2(x/2) \, (\sqrt{2} + \sqrt{1 + \cos x})} = \frac{2}{4\cos^2(x/2) \, (\sqrt{2} + \sqrt{1 + \cos x})}

Cancel $\sin^2(x/2)$ (valid for $x \neq 0$ , and the limit as $x \to 0$ is unaffected). This simplifies to:

$\frac{1}{2\cos^2(x/2) \, (\sqrt{2} + \sqrt{1 + \cos x})}$ …

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