$\lim_{x \to 0} \dfrac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}}$ is (A) $2$ (B) $0$ (C) $1$ (D) $-1$

limx → 0 dfracsin x√(x + 1) - √(1 - x) is (A) 2 (B – Class 11 Mathematics

NCERT Exemplar · Q60

\lim_{x \to 0} \dfrac{\sin x}{\sqrt{x + 1} - \sqrt{1 - 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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The problem asks for the limit of an expression that results in an indeterminate form. The key idea is to resolve this $\frac{0}{0}$ form using rationalization and applying the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

As $x \to 0$ , the numerator $\sin x \to 0$ and the denominator $\sqrt{0+1} - \sqrt{1-0} = 1 - 1 = 0$ . This is an indeterminate form $\frac{0}{0}$ .
To resolve the square roots in the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x + 1} + \sqrt{1 - x}$ :

\lim_{x \to 0} \dfrac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}} \times \dfrac{\sqrt{x + 1} + \sqrt{1 - x}}{\sqrt{x + 1} + \sqrt{1 - x}}

Simplify the denominator using $(a-b)(a+b) = a^2 - b^2$ :

(\sqrt{x + 1})^2 - (\sqrt{1 - x})^2 = (x + 1) - (1 - x) = x + 1 - 1 + x = 2x

The expression becomes:
$$ \lim_{x \to 0} \dfrac{\sin x (\sqrt{x + 1} + \sqrt{1 - x})}{2x} $$ …

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The limit is of the $\frac{0}{0}$ indeterminate form. Rationalizing the denominator and using the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ simplifies the expression to $\lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{\sqrt{x+1} + \sqrt{1-x}}{2}$ , which evaluates to $\boxed{1}$ .

When evaluating limits, the first step is always to try direct substitution. If we substitute $x=0$ into the given expression:

Numerator: $\sin(0) = 0$

Denominator: $\sqrt{0+1} - \sqrt{1-0} = \sqrt{1} - \sqrt{1} = 1 - 1 = 0$

Since we obtain the indeterminate form $\frac{0}{0}$ , direct substitution does not give us the limit. This indicates that there's a common factor in the numerator and denominator that approaches zero as $x \to 0$ , which needs to be cancelled out.

The presence of square roots in the denominator, specifically in the form $\sqrt{A} - \sqrt{B}$ , is a strong hint to use the technique of rationalization. By multiplying the numerator and denominator by the conjugate of the denominator, $(\sqrt{x+1} + \sqrt{1-x})$ , we can eliminate the square roots from the denominator using the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$ . This algebraic manipulation often simplifies the expression, allowing us to cancel the problematic term.

Additionally, the numerator contains $\sin x$ . This immediately brings to mind a fundamental trigonometric limit:

\lim_{x \to 0} \frac{\sin x}{x} = 1

Watch out

While L'Hôpital's Rule can be applied to $\frac{0}{0}$ forms, it's often more complex with square roots. Mastering algebraic simplification and standard limits is crucial, as these methods are often more direct and less prone to calculation errors.

Let's proceed with the solution:

Identify the indeterminate form:

As shown above, substituting $x=0$ into the expression yields $\frac{\sin 0}{\sqrt{0+1} - \sqrt{1-0}} = \frac{0}{1-1} = \frac{0}{0}$ . This confirms we need to simplify the expression.
Rationalize the denominator:

To remove the square roots from the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{x+1} + \sqrt{1-x}$ .

\lim_{x \to 0} \dfrac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}} = \lim_{x \to 0} \dfrac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}} \cdot \dfrac{\sqrt{x + 1} + \sqrt{1 - x}}{\sqrt{x + 1} + \sqrt{1 - x}}

Simplify the expression: Apply the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$ , to the denominator. Here, $a = \sqrt{x+1}$ and $b = \sqrt{1-x}$ . The denominator becomes: …

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Q.lim⁡x→0sin⁡xx+1−1−x\lim_{x \to 0} \dfrac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}}limx→0​x+1​−1−x​sinx​ is (A) 222 (B) 000 (C) 111 (D) −1-1−1

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Q. $\lim_{x \to 0} \dfrac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}}$ is
(A) $2$
(B) $0$
(C) $1$
(D) $-1$

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