📝 Indeterminate form 0/0 limits (16 MCQs)
📖 From Calculus • 2. Limits and Continuity an Introduction • 16 questions available
What is Indeterminate form 0/0 limits?
Definition:
An indeterminate form of type occurs when evaluating yields , meaning both numerator and denominator approach zero, providing no immediate information about the limit. Such limits require algebraic techniques like factorization, rationalization, or trigonometric identities to simplify the expression and cancel the common factor causing the zero, after which the limit can be evaluated.
Example:
Evaluate .
Solution: form; factor , then .
Reason:
This form is critical because it arises in derivative definitions and many real-world problems, and mastering its resolution is key to understanding how to manipulate functions to reveal their true limiting behavior.
📝 All Indeterminate form 0/0 limits MCQs
Q1. What is the limit \\\displaystyle \\lim_{x\\to 2}\\frac{x^{2}-4}{x-2}\?
📖 Explanation: Factor the numerator as \(x-2)(x+2)\; the factor \(x-2)\ cancels with the denominator, leaving \x+2\. Substituting \x=2\ gives \2+2=4\. Thus the expression approaches 4, so the limit equals 4. The cancellation is valid for values of \x\ near but not equal to 2.
Q2. If \\\lim_{x\\to a}f(x)=L\ and \\\lim_{x\\to a}g(x)=0\ where both functions are continuous near \a\, which statement must be true about \\\displaystyle \\lim_{x\\to a}\\frac{f(x)}{g(x)}\?
📖 Explanation: Because the denominator approaches zero while the numerator approaches a non‑zero constant, the quotient can blow up, approach a finite number, or oscillate, depending on how fast each function approaches its limit. Without additional information, no single value is forced, so any real number is possible.
Q3. Evaluate the limit \\\displaystyle \\lim_{x\\to 0}\\frac{e^{x}-1}{x}\.
📖 Explanation: Using the series expansion \e^{x}=1+x+\\frac{x^{2}}{2}+\\dots\, subtract 1 to obtain \x+\\frac{x^{2}}{2}+\\dots\. Dividing by \x\ yields \1+\\frac{x}{2}+\\dots\, which approaches 1 as \x\\to0\. Hence the limit equals 1.
Q4. Find \\\displaystyle \\lim_{x\\to 0}\\frac{\\ln(1+x)}{x}\.
📖 Explanation: Apply L'Hôpital's Rule because the expression is \0/0\. Differentiate numerator and denominator: \\\frac{d}{dx}\\ln(1+x)=\\frac{1}{1+x}\ and \\\frac{d}{dx}x=1\. The new limit is \\\lim_{x\\to0}\\frac{1}{1+x}=1\. Therefore the original limit equals 1.
Q5. What is \\\displaystyle \\lim_{x\\to 0^{+}}\\frac{|x|}{x}\?
📖 Explanation: For positive \x\, \|x|=x\. Hence the quotient \\\frac{|x|}{x}=\\frac{x}{x}=1\ for all \x>0\. As \x\ approaches 0 from the right, the value remains 1, so the one‑sided limit is 1.
Q6. Compare the limits \\\displaystyle \\lim_{x\\to 3}\\frac{x^{2}-9}{x-3}\ and \\\displaystyle \\lim_{x\\to 3}\\frac{x^{3}-27}{x-3}\. Which statement is correct?
📖 Explanation: Factor each numerator: \x^{2}-9=(x-3)(x+3)\ gives limit \x+3\\to6\. For \x^{3}-27=(x-3)(x^{2}+3x+9)\ the limit is \x^{2}+3x+9\\to27\. Since 27>6, the second limit is larger.
Q7. Evaluate \\\displaystyle \\lim_{x\\to 0}\\frac{1-\\cos x}{x^{2}}\ and select its value.
📖 Explanation: Rewrite \1-\\cos x\ as \2\\sin^{2}\\frac{x}{2}\. The expression becomes \\\frac{2\\sin^{2}(x/2)}{x^{2}}=\\frac{2}{4}\\left(\\frac{\\sin(x/2)}{x/2}\\right)^{2}\. As \x\\to0\, the sine ratio approaches 1, giving \\\frac{1}{2}\.
Q8. Given \f(x)=x^{2}\\sin\\frac{1}{x}\ for \x\\neq0\ and \f(0)=0\, what is \\\displaystyle \\lim_{x\\to 0}\\frac{f(x)}{x}\?
📖 Explanation: The quotient simplifies to \x\\sin\\frac{1}{x}\. Since \|\\sin\\frac{1}{x}|\\le1\, we have \|x\\sin\\frac{1}{x}|\\le|x|\. As \x\\to0\, the bound forces the expression to 0, so the limit is 0.
Q9. Find \\\displaystyle \\lim_{x\\to 1}\\frac{x^{3}-1}{x-1}\.
📖 Explanation: Factor the numerator: \x^{3}-1=(x-1)(x^{2}+x+1)\. Cancel \x-1\ to obtain \x^{2}+x+1\. Substituting \x=1\ yields \1+1+1=3\. Hence the limit equals 3.
Q10. What is \\\displaystyle \\lim_{x\\to 0}\\frac{x-\\sin x}{x^{3}}\?
📖 Explanation: Use the series \\\sin x = x-\\frac{x^{3}}{6}+\\dots\. Subtracting from \x\ gives \\\frac{x^{3}}{6}+\\dots\. Dividing by \x^{3}\ leaves \\\frac{1}{6}+\\dots\, which approaches \\\frac{1}{6}\ as \x\\to0\.
Q11. Using the identity \1-\\cos x = 2\\sin^{2}\\frac{x}{2}\, evaluate \\\displaystyle \\lim_{x\\to 0}\\frac{1-\\cos x}{x^{2}}\.
📖 Explanation: Replace the numerator: \\\frac{2\\sin^{2}(x/2)}{x^{2}}=\\frac{2}{4}\\left(\\frac{\\sin(x/2)}{x/2}\\right)^{2}\. The sine ratio tends to 1 as \x\\to0\, giving \\\frac{2}{4}=\\frac{1}{2}\. Thus the limit is \\\frac{1}{2}\.
Q12. Which limit exhibits the \0/0\ indeterminate form but cannot be resolved by simple factoring, thus requiring L'Hôpital's Rule?
📖 Explanation: The first, second, and fourth limits can be simplified by factoring the numerator, removing the zero denominator. The limit \\\frac{\\ln x}{x-1}\ as \x\\to1\ yields \0/0\ but does not factor nicely; applying L'Hôpital's Rule differentiates numerator and denominator, giving \\\frac{1/x}{1}=1\.
Q13. Which statement correctly defines an indeterminate form of type \0/0\?
📖 Explanation: An indeterminate form \0/0\ occurs when the limit of a fraction has both its numerator and denominator tending to zero. This situation does not determine the limit's value without further analysis, such as factoring, applying L'Hôpital's Rule, or using series expansions.
Q14. Determine \\\displaystyle \\lim_{x\\to 0}\\frac{\\tan x - x}{x^{3}}\ using series expansion.
📖 Explanation: Expand \\\tan x\ as \x+\\frac{x^{3}}{3}+\\frac{2x^{5}}{15}+\\dots\. Subtracting \x\ leaves \\\frac{x^{3}}{3}+\\dots\. Dividing by \x^{3}\ gives \\\frac{1}{3}+\\dots\, which approaches \\\frac{1}{3}\ as \x\\to0\. The limit differs from the similar sine limit because the cubic term coefficient in the tangent series is \1/3\ rather than \1/6\.
Q15. If \\\displaystyle \\lim_{x\\to a}h(x)=L\\neq0\ and \\\displaystyle \\lim_{x\\to a}k(x)=0\, which statement about \\\displaystyle \\lim_{x\\to a}\\frac{h(x)}{k(x)}\ is always true?
📖 Explanation: When the denominator tends to zero while the numerator approaches a non‑zero finite number, the quotient can blow up to \+\\infty\ or \-\\infty\, converge to a finite value, or oscillate, depending on how quickly each function approaches its limit. Therefore no single outcome is guaranteed; the result depends on the functions' rates.
Q16. What is \\\displaystyle \\lim_{x\\to 0}\\frac{x}{x}\?
📖 Explanation: For all \x\\neq0\, the expression simplifies to \1\. As \x\ approaches 0, the value remains 1, so the limit exists and equals 1. The original form \0/0\ is removable by cancellation, leading to the constant function 1.