Evaluate limx → π dfrac1 - sin(x)/(2)cos(x)/(2)(co – Class 11 Mathematics

NCERT Exemplar · Q50

\lim_{x \to \pi} \dfrac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)}

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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 limit is of the form $\frac{0}{0}$ ; we simplify using trigonometric identities and a substitution $t = \frac{x}{4}$ , then cancel the common factor to get the value $\frac{1}{\sqrt{2}}$ .

We are evaluating

\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)}.

Direct substitution of $x = \pi$ gives $\frac{1 - \sin(\pi/2)}{\cos(\pi/2)(\cos(\pi/4) - \sin(\pi/4))} = \frac{1 - 1}{0 \cdot (\text{nonzero})} = \frac{0}{0}$ , an indeterminate form. So we need to simplify algebraically.

The key idea: rewrite the numerator $1 - \sin\frac{x}{2}$ using a double-angle identity, and see if it cancels with part of the denominator.

Rewrite the numerator Recall $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ . Here $\theta = \frac{x}{2}$ , so

\sin\frac{x}{2} = 2\sin\frac{x}{4}\cos\frac{x}{4}.

Hence

1 - \sin\frac{x}{2} = 1 - 2\sin\frac{x}{4}\cos\frac{x}{4}.

This looks like $1 - \sin 2u$ with $u = \frac{x}{4}$ , but we can also use the identity $1 - \sin 2u = (\cos u - \sin u)^2$ . Let's verify:

(\cos u - \sin u)^2 = \cos^2 u + \sin^2 u - 2\sin u\cos u = 1 - \sin 2u.

Perfect. So with $u = \frac{x}{4}$ ,

1 - \sin\frac{x}{2} = \left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)^2.

Substitute into the limit The denominator already contains $\cos\frac{x}{2}\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)$ . So the expression becomes

\frac{\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)^2}{\cos\frac{x}{2}\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)}.

Cancel one factor of $\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)$ , provided it is not zero at $x = \pi$ (it is not: at $x=\pi$ , $\cos\frac{\pi}{4} - \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$ — careful! Actually it is zero. So cancellation is valid only for $x \neq \pi$ , but that's fine for a limit; we are looking at values near $\pi$ , not at $\pi$ itself. The factor cancels, leaving a simpler expression that is continuous at $x=\pi$ after further simplification.)

After cancellation we get

\frac{\cos\frac{x}{4} - \sin\frac{x}{4}}{\cos\frac{x}{2}}.

Simplify the denominator

Use the double-angle identity for cosine: $\cos\frac{x}{2} = \cos\left(2\cdot\frac{x}{4}\right) = \cos^2\frac{x}{4} - \sin^2\frac{x}{4} = (\cos\frac{x}{4} - \sin\frac{x}{4})(\cos\frac{x}{4} + \sin\frac{x}{4})$ .

So the limit becomes

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Q.Evaluate $\lim_{x \to \pi} \dfrac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)}$ .

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