$\lim_{x \to 0} \left( \sin mx \cot \dfrac{x}{\sqrt{3}} \right) = 2$, then $m = $ ________.

limx → 0 ( sin mx cot dfracx√(3) ) = 2, then m = ____.

NCERT Exemplar · Q78

\lim_{x \to 0} \left( \sin mx \cot \dfrac{x}{\sqrt{3}} \right) = 2

Sikkim CbseShort· 3mImportance★★★★★est

99% · 173/175 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

The key idea here is the Limit of Trigonometric Functions, specifically using the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$ .

We are given the limit:

\lim_{x \to 0} \left( \sin mx \cot \dfrac{x}{\sqrt{3}} \right) = 2

Rewrite the cotangent term in terms of sine and cosine:

\lim_{x \to 0} \left( \sin mx \cdot \dfrac{\cos \dfrac{x}{\sqrt{3}}}{\sin \dfrac{x}{\sqrt{3}}} \right)

Rearrange the expression to apply the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$ :

\lim_{x \to 0} \left( \dfrac{\sin mx}{mx} \cdot mx \cdot \dfrac{\frac{x}{\sqrt{3}}}{\sin \dfrac{x}{\sqrt{3}}} \cdot \dfrac{1}{\frac{x}{\sqrt{3}}} \cdot \cos \dfrac{x}{\sqrt{3}} \right)

This simplifies to:

\lim_{x \to 0} \left( \dfrac{\sin mx}{mx} \cdot \dfrac{\frac{x}{\sqrt{3}}}{\sin \dfrac{x}{\sqrt{3}}} \cdot \dfrac{mx}{\frac{x}{\sqrt{3}}} \cdot \cos \dfrac{x}{\sqrt{3}} \right)

Evaluate each part of the limit as $x \to 0$ $x \to 0$ :
- $\lim_{x \to 0} \dfrac{\sin mx}{mx} = 1$ …

Start your 14-day free trial to unlock the full solution →

The limit involves an indeterminate form $0 \cdot \infty$ . By rewriting $\cot x$ as $1/\tan x$ and using the standard limits $\lim_{x \to 0} \frac{\sin kx}{kx} = 1$ and $\lim_{x \to 0} \frac{\tan kx}{kx} = 1$ , we simplify the expression to solve for $m$ . The value of $m$ is $\boxed{2/\sqrt{3}}$ .

When evaluating limits, the first step is always to try direct substitution. If we substitute $x=0$ into the given expression $\sin mx \cot \frac{x}{\sqrt{3}}$ , we get:

$\sin(m \cdot 0) \cot \left( \frac{0}{\sqrt{3}} \right) = \sin(0) \cot(0)$

We know $\sin(0) = 0$ . However, $\cot(0)$ is undefined, as $\cot(0) = \frac{\cos(0)}{\sin(0)} = \frac{1}{0}$ , which tends to infinity. So, we have an indeterminate form of type $0 \cdot \infty$ . This means we need to manipulate the expression algebraically to resolve the indeterminacy before evaluating the limit.

The key idea here is to transform the expression into forms where we can apply standard trigonometric limits, specifically those involving $\frac{\sin \theta}{\theta}$ and $\frac{\tan \theta}{\theta}$ as $\theta \to 0$ .

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

We will rewrite $\cot \theta$ as $\frac{1}{\tan \theta}$ to make use of the second standard limit.

Let's evaluate the limit step-by-step.

Rewrite the expression using $\cot \theta = \frac{1}{\tan \theta}$ . The given limit is $\lim_{x \to 0} \left( \sin mx \cot \dfrac{x}{\sqrt{3}} \right)$ . We can rewrite this as:

\lim_{x \to 0} \left( \sin mx \cdot \frac{1}{\tan \frac{x}{\sqrt{3}}} \right) = \lim_{x \to 0} \frac{\sin mx}{\tan \frac{x}{\sqrt{3}}}

Now, if we substitute $x=0$, we get $\frac{\sin 0}{\tan 0} = \frac{0}{0}$, which is another indeterminate form. This confirms our need for further manipulation.

2. Introduce terms to match standard limit forms.

To use the standard limits $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ , we need to divide $\sin mx$ by $mx$ and $\tan \frac{x}{\sqrt{3}}$ by $\frac{x}{\sqrt{3}}$ . To maintain the equality, we must also multiply by these terms.

\lim_{x \to 0} \frac{\sin mx}{\tan \frac{x}{\sqrt{3}}} = \lim_{x \to 0} \frac{\left( \frac{\sin mx}{mx} \right) \cdot mx}{\left( \frac{\tan \frac{x}{\sqrt{3}}}{\frac{x}{\sqrt{3}}} \right) \cdot \frac{x}{\sqrt{3}}}

Apply the standard limits. As $x \to 0$ , we have $mx \to 0$ and $\frac{x}{\sqrt{3}} \to 0$ . Therefore, we can apply the standard limits:

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

\lim_{x \to 0} \frac{\tan \frac{x}{\sqrt{3}}}{\frac{x}{\sqrt{3}}} = 1

Substituting these values into our expression:
$$ \lim_{x \to 0} \frac{\left( \frac{\sin mx}{mx} \right) \cdot mx}{\left( \frac{\tan \frac{x}{\sqrt{3}}}{\frac{x}{\sqrt{3}}} \right) \cdot \frac{x}{\sqrt{3}}} = \frac{1 \cdot mx}{1 \cdot \frac{x}{\sqrt{3}}} $$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ77 — If f(x) = (tan x)/(x - π), then limx → π f(x) = ____.→Next questionQ79 — If y = 1 + (x)/(1!) + (x²)/(2!) + (x³)/(3!) + ..., then (dy)/(dx) = __…

Q.lim⁡x→0(sin⁡mxcot⁡x3)=2\lim_{x \to 0} \left( \sin mx \cot \dfrac{x}{\sqrt{3}} \right) = 2limx→0​(sinmxcot3​x​)=2, then m=m = m= ________.

🎓Unlock everything free for 14 days

Q. $\lim_{x \to 0} \left( \sin mx \cot \dfrac{x}{\sqrt{3}} \right) = 2$ , then $m =$ ________.

Unlock everything free for 14 days