Limits of transcendental functions

Why $0/0$ needs a new idea. Direct substitution works for polynomials and most rational functions, but $\dfrac{\sin x}{x}$ at $x=0$ gives $\tfrac{0}{0}$ — an indeterminate form that carries no information on its own. The ratio does settle to a definite number, but finding it needs a genuinely new limit, not factorisation.

The standard result, accepted and used at this level (a geometric sketch using areas of a triangle and a circular sector squeezed together motivates it, but no rigorous proof is demanded):

where: $x$ must be measured in radians — the result is false, or at least needs an extra conversion factor, in degrees.

Two immediate consequences, combining this with the algebra-of-limits theorems (limit of a product/quotient equals product/quotient of limits, when the pieces exist):

where: the second equality in the first result uses direct substitution of $\cos x$ at $x=0$, itself justified because $\cos x$ is well-behaved (continuous-looking) at this level.

Away from zero, use direct substitution. Since $\sin x$, $\cos x$ and $\tan x$ (where defined) behave smoothly at any point $a$ in this course:

Where this sits in the syllabus. This continues the "Calculus (Limits and Derivatives)" chapter of ISC Class 11 Mathematics, immediately after limits of algebraic functions, and leans on trigonometric identities from the "Trigonometry" sub-topic of "Sets and Functions." Inverse trigonometric functions and any technique that presumes derivatives (such as L'Hôpital's rule) are outside this chapter's scope — every limit here must be cracked with algebra, substitution and identities alone.

Why the $0/0$ form is special, not fatal. An indeterminate form such as $\tfrac{0}{0}$ does not mean "the limit does not exist" — it means the expression, written that particular way, hides the answer. The whole discipline of this chapter is learning a small toolkit of rewrites (dividing by $x$, using a double-angle identity, splitting a product) that turns a hidden $0/0$ into a form where every piece has a known limit, after which the algebra-of-limits theorems (sum, product, quotient of limits equal the limit of the sum, product, quotient, provided each piece's limit exists) finish the job.

Reshaping harder limits. More elaborate expressions are handled by combining algebra-of-limits with trigonometric identities (angle addition, double-angle formulae) to reduce the expression to a product of pieces of the standard $\dfrac{\sin(\text{something}\to 0)}{\text{something}\to 0}$ shape. Substituting $x=k\theta$ gives a useful generalised pair:

where: $k$, $a$, $b$ are fixed real constants and $\theta\to 0$.

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin 5x}{\sin 3x}$.

Step 1. Try direct substitution. At $x=0$: numerator $=\sin 0=0$, denominator $=\sin 0=0$ — this is $\tfrac{0}{0}$, so substitution fails and a rewrite is needed.

Step 2. Divide numerator and denominator by $x$ to expose the standard shape in each piece:

Step 3. As $x\to 0$, both $5x\to 0$ and $3x\to 0$, so the standard limit applies to each factor: $\lim_{x\to 0}\dfrac{\sin 5x}{5x}=1$ and $\lim_{x\to 0}\dfrac{\sin 3x}{3x}=1$.

Step 4. Combine using the quotient rule for limits: $\lim_{x\to 0}\dfrac{\sin 5x}{\sin 3x} = \dfrac{5\cdot 1}{3\cdot 1} = \dfrac{5}{3}$.

The final answer is $\boxed{5/3}$.

Competitive stretch

The same trick generalises instantly: for any constants $a,b$ with $b\neq 0$, $\lim_{x\to 0}\dfrac{\sin ax}{\sin bx} = \dfrac{a}{b}$, obtained by the identical divide-by-$x$ step. Recognising this pattern lets you write down the answer to $\lim_{x\to 0}\sin 7x/\sin 2x = 7/2$ without redoing the algebra each time. The same idea extends to tangent: since $\tan x/x\to 1$ as well, a ratio like $\lim_{x\to 0}\dfrac{\tan 4x}{\sin 2x}$ can be split into $\dfrac{\tan 4x}{4x}\cdot\dfrac{2x}{\sin 2x}\cdot\dfrac{4x}{2x}$, each of the first two factors tending to $1$ and the last to $2$, giving $2$ — the general lesson being that any ratio built purely from $\sin$ and $\tan$ of linear arguments in $x$ reduces, in the limit, to the ratio of those linear coefficients.

• Applying $\lim_{x\to 0}\sin x/x=1$ when the angle inside the sine and the denominator do not match, without first multiplying/dividing by the right constant to align them (as in the worked example).
• Using the standard limit for angles measured in degrees — it holds only in radians; a degree version needs an extra conversion factor and is not the expected form here.
• Reaching for derivative-based shortcuts such as L'Hôpital's rule — this chapter deliberately expects limits evaluated by algebraic and trigonometric manipulation, since derivatives have not yet been introduced.
• Forgetting the identity (double-angle, sum-to-product) that would convert an unfamiliar expression into the standard $\sin(\cdot)/(\cdot)$ shape, and giving up instead of rewriting.
• Skipping the direct-substitution check at the start — many limits at points other than zero are simple substitutions, and reaching straight for identities wastes time and invites algebra mistakes on a problem that needed none.

Ask the tutor: If a trigonometric limit does not obviously reduce to the standard $\sin x/x$ form, ask your AI tutor which identity or algebraic split to try first.