CUET UG Mathematics Calculus Limits and Continuity LHôpitals Rule Sandwich Theorem

Must‑Know (15–20 detailed bullets)

• L'Hôpital's Rule applies only when the limit is of the indeterminate form ( \frac{0}{0} ) or ( \frac{\infty}{\infty} ). Example: ( \lim_{x \to 0} \frac{\sin x}{x} = \frac{0}{0} ), so apply L'Hôpital: ( \lim_{x \to 0} \frac{\cos x}{1} = 1 ).

• To use L'Hôpital's Rule, differentiate numerator and denominator separately; do not use quotient rule. Example: ( \lim_{x \to 0} \frac{1 - \cos x}{x^2} ) → differentiate to get ( \lim_{x \to 0} \frac{\sin x}{2x} ), still ( \frac{0}{0} ), apply again: ( \lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2} ).

• Indeterminate forms like ( 0 \cdot \infty ), ( \infty - \infty ), ( 1^\infty ), ( 0^0 ), ( \infty^0 ) must be converted to ( \frac{0}{0} ) or ( \frac{\infty}{\infty} ) before applying L'Hôpital. Example: ( \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} = \frac{-\infty}{\infty} ), apply L'Hôpital: ( \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0 ).

• The Sandwich (Squeeze) Theorem states: if ( f(x) \leq g(x) \leq h(x) ) for all ( x ) near ( a ) (except possibly at ( a )), and ( \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L ), then ( \lim_{x \to a} g(x) = L ).

• A standard application of the Sandwich Theorem is ( \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0 ), because ( -|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x| ), and both bounds go to 0.

• ( \lim_{x \to 0} \frac{\tan x}{x} = 1 ) — this is derived using ( \frac{\sin x}{x} \to 1 ) and ( \cos x \to 1 ), or via L'Hôpital: ( \lim_{x \to 0} \frac{\sec^2 x}{1} = 1 ).

• ( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 ); if encountered as ( \frac{0}{0} ), L'Hôpital gives ( \lim_{x \to 0} \frac{e^x}{1} = 1 ).

• ( \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1 ); applying L'Hôpital: ( \lim_{x \to 0} \frac{1/(1+x)}{1} = 1 ).

• L'Hôpital’s Rule can be applied repeatedly if the form remains indeterminate after first differentiation. Example: ( \lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \frac{0}{0} ), first derivative: ( \frac{e^x - 1}{2x} = \frac{0}{0} ), second: ( \frac{e^x}{2} = \frac{1}{2} ).

• The limit ( \lim_{x \to \infty} \frac{\ln x}{x} = 0 ); use L'Hôpital: ( \lim_{x \to \infty} \frac{1/x}{1} = 0 ).

• ( \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 ); direct substitution gives ( \frac{0}{0} ), L'Hôpital: ( \lim_{x \to 0} \frac{\sin x}{1} = 0 ).

• For ( \lim_{x \to a} \frac{f(x)}{g(x)} ), if ( \lim f(x) = 0 ), ( \lim g(x) = 5 ), then limit is 0 — not indeterminate, so L'Hôpital does not apply.

• The function ( f(x) = \begin{cases} x^2 \sin(1/x), & x \ne 0 \ 0, & x = 0 \end{cases} ) is continuous at 0 because ( \lim_{x \to 0} x^2 \sin(1/x) = 0 ) by Sandwich Theorem.

• ( \lim_{x \to 0} \frac{\sin 3x}{x} = 3 ), since ( \frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x} \to 3 \cdot 1 = 3 ).

• If ( \lim_{x \to a} f(x) = \infty ), ( \lim_{x \to a} g(x) = \infty ), then ( \lim_{x \to a} \frac{f(x)}{g(x)} ) may still be finite — L'Hôpital may help. Example: ( \lim_{x \to \infty} \frac{x}{e^x} = 0 ) (via L'Hôpital).

• The expression ( \lim_{x \to 0} \frac{\cos x}{x} ) is not indeterminate; it diverges to ( \infty ) (from right) or ( -\infty ) (from left), so L'Hôpital does not apply.

• ( \lim_{x \to 0^+} x^x = 1 ); rewrite as ( e^{\lim_{x \to 0^+} x \ln x} = e^0 = 1 ), using earlier result ( x \ln x \to 0 ).

• ( \lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1 ); apply L'Hôpital: ( \lim_{x \to 0} \frac{1/(1+x^2)}{1} = 1 ).

• The Sandwich Theorem is especially useful for limits involving trigonometric functions with oscillating behavior near 0.

• Verify from NCERT: exact number of solved examples on L'Hôpital’s Rule in Class 12 NCERT Chapter 5 (Continuity and Differentiability).

Difficulty Level

Intermediate — because L'Hôpital’s Rule requires recognizing indeterminate forms and correct differentiation, while Sandwich Theorem needs construction of bounding functions, both frequently tested in multi-step problems.

Common CUET Traps

• Trap: Applying L'Hôpital’s Rule to limits that are not in ( \frac{0}{0} ) or ( \frac{\infty}{\infty} ) form, such as ( \frac{1}{0} ) or ( \frac{2}{3} ).
  Avoid: Always check the form first; if it's determinate, evaluate directly.

• Trap: Differentiating the entire fraction using quotient rule instead of differentiating numerator and denominator separately.
  Avoid: L'Hôpital’s Rule requires ( \lim \frac{f'(x)}{g'(x)} ), not derivative of ( \frac{f(x)}{g(x)} ).

• Trap: Assuming ( \lim_{x \to 0} \frac{\sin x}{x} = 0 ) because numerator is 0 — ignoring the indeterminate form.
  Avoid: Remember standard limit: ( \lim_{x \to 0} \frac{\sin x}{x} = 1 ), proven via geometric argument or L'Hôpital.

Practice MCQs

Q1. What is ( \lim_{x \to 0} \frac{\sin 5x}{x} )?
A. 0
B. 1
C. 5
D. ( \frac{1}{5} )
Answer: C
Explanation: ( \frac{\sin 5x}{x} = 5 \cdot \frac{\sin 5x}{5x} \to 5 \cdot 1 = 5 ).
Why others fail: Option B is tempting if student forgets to scale by 5.

Q2. Which of the following is an indeterminate form to which L'Hôpital’s Rule can be applied?
A. ( \frac{1}{0} )
B. ( \frac{0}{5} )
C. ( \frac{\infty}{\infty} )
D. ( \frac{2}{3} )
Answer: C
Explanation: ( \frac{\infty}{\infty} ) is a valid indeterminate form for L'Hôpital.
Why others fail: Option A looks like ∞ but is not indeterminate — it diverges.

Q3. Evaluate ( \lim_{x \to 0} \frac{e^x - 1 - x}{x^2} ).
A. 0
B. ( \frac{1}{2} )
C. 1
D. ∞
Answer: B
Explanation: Apply L'Hôpital twice: first get ( \frac{e^x - 1}{2x} ), then ( \frac{e^x}{2} \to \frac{1}{2} ).
Why others fail: Option A is tempting if student stops after first application and substitutes incorrectly.

Q4. Using the Sandwich Theorem, what is ( \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) )?
A. 0
B. 1
C. undefined
D. ( \frac{1}{2} )
Answer: A
Explanation: Since ( -1 \leq \cos(1/x) \leq 1 ), so ( -x^2 \leq x^2 \cos(1/x) \leq x^2 ), both bounds → 0.
Why others fail: Option C is tempting if student thinks oscillation implies no limit.

Q5. ( \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} ) equals:
A. 0
B. 1
C. ∞
D. ( \frac{1}{2} )
Answer: A
Explanation: Apply L'Hôpital: ( \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0 ).
Why others fail: Option D is tempting if student incorrectly differentiates denominator.

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ L'Hôpital’s Rule only for ( \frac{0}{0} ) or ( \frac{\infty}{\infty} ) — not for ( \frac{0}{\infty} ) or ( \frac{\infty}{0} ).
• ⚠️ ( \lim_{x \to 0} \frac{\sin x}{x} = 1 ) — not 0 or ∞.
• ⚠️ ( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} ) — remember this standard result.
• ⚠️ Never apply L'Hôpital to sum/difference — convert to quotient first if needed.
• ⚠️ Sandwich Theorem requires both upper and lower bounds to converge to same limit.
• ⚠️ ( \lim_{x \to 0} x \sin(1/x) = 0 ) — classic Sandwich example.
• ⚠️ ( \lim_{x \to \infty} \frac{\log x}{x^n} = 0 ) for any ( n > 0 ).
• ⚠️ ( \lim_{x \to 0} \frac{\tan x}{x} = 1 ) — same as ( \frac{\sin x}{x} ).
• ⚠️ ( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 ) — exponential standard limit.
• ⚠️ ( \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1 ) — logarithmic standard limit.
• ⚠️ If limit is ( 0 \cdot \infty ), rewrite as ( \frac{0}{1/\infty} = \frac{0}{0} ) or ( \frac{\infty}{1/0} = \frac{\infty}{\infty} ).
• ⚠️ L'Hôpital can be applied multiple times — until form becomes determinate.
• ⚠️ ( \lim_{x \to 0^+} x \ln x = 0 ) — use L'Hôpital after rewriting.
• ⚠️ ( \lim_{x \to 0} \frac{\cos x - 1}{x} = 0 ) — apply L'Hôpital or use identity.
• ⚠️ Oscillating functions (like ( \sin(1/x) )) can have limits if bounded by functions going to 0.
• ⚠️ ( \lim_{x \to a} \frac{f(x)}{g(x)} ) does not exist if ( f \to L \ne 0 ), ( g \to 0 ) — infinite limit, not indeterminate.
• ⚠️ Always verify differentiability before applying L'Hôpital — required in theory.
• ⚠️ ( \lim_{x \to 0} (1 + x)^{1/x} = e ) — not directly L'Hôpital, but related via log.
• ⚠️ Mnemonic: “L’Hospital” — think “limits in hospital” when form is sick (indeterminate).
• ⚠️ Verify from NCERT: L'Hôpital’s Rule is not explicitly named in NCERT but applications appear in solved examples.