Questions: Limits and Derivatives

Limits and Derivatives24 min readLesson 6 of 7

in hinglish

Chapter ka quick recap. Limit ke liye left aur right dono sides dekho; zero by zero aaye to factor, rationalise ya standard limit lagao. Trigonometric limits me sin x/x \to 1, (1-cos x)/x² \to 1/2 yaad rakho, aur radians me hi kaam karo. Sandwich theorem oscillation wale questions me kaam aata hai. Derivative first principles se [f(x+h)-f(x)]/h ka limit hota hai; power rule, sum rule aur product rule se calculation fast hoti hai. Standard derivatives: sin\tocos, cos\to-sin, tan\tosec². Neeche das sawal hain; har sawal me pehle expression simplify karo, phir limit ya derivative lo. Answer ko rough check se verify karo. Derivative nikaalne ke baad ek baar sochein ki answer ka shape sahi hai ya nahi: polynomial ki degree ek kam honi chahiye aur product rule me dono terms hone chahiye.

Q1. Find the limit of (x squared minus 4) by (x minus 2) as x tends to 2. Answer: Substitution gives zero over zero, so factor the numerator as (x minus 2)(x plus 2). Cancelling the common factor leaves x plus 2, whose limit at x equal to 2 is 4. Hinglish: Zero by zero form hai; (x²-4) = (x-2)(x+2), cancel karke limit 4.

Q2. Evaluate the limit of sine 3x by x as x tends to 0. Answer: Write sine 3x by x as 3 times (sine 3x by 3x). The bracket tends to 1, so the limit is 3. Hinglish: sin 3x/x = 3\cdot(sin 3x/3x) \to 3\cdot1 = 3.

Q3. Find the limit of (1 minus cosine x) by x squared as x tends to 0. Answer: This is the standard limit with value 1 by 2. It can be checked by multiplying numerator and denominator by (1 plus cosine x), which turns the numerator into sine squared x and the denominator into x squared times (1 plus cosine x), giving 1 times 1 by 2. Hinglish: Standard limit (1-cos x)/x² = 1/2; conjugate se bhi wahi aata hai.

Q4. Find the limit of x times sine of 1 by x as x tends to 0. Answer: Since sine of 1 by x lies between minus 1 and 1, the expression lies between minus the absolute value of x and plus the absolute value of x. Both bounds tend to zero, so by the sandwich theorem the limit is zero. Hinglish: |sin(1/x)| \leq 1, isliye x\cdotsin(1/x) ka limit 0 hai (sandwich theorem).

Q5. Differentiate x cubed plus 2x squared minus 5x plus 7. Answer: Apply the power rule term by term: 3x squared plus 4x minus 5. The constant 7 differentiates to zero. The degree drops from 3 to 2, as expected. Hinglish: Power rule se 3x² + 4x - 5; constant ka derivative zero.

Q6. Find the derivative of x sine x. Answer: Use the product rule: the derivative of x times sine x plus x times the derivative of sine x, which is sine x plus x cosine x. Hinglish: Product rule: (x)'\cdotsin x + x\cdot(sin x)' = sin x + x cos x.

Q7. Find the derivative of cosine x from first principles. Answer: The difference quotient is (cosine of x plus h minus cosine x) by h. Using the cosine addition formula gives cosine x times (cosine h minus 1) by h minus sine x times (sine h) by h. The two standard limits give 0 minus sine x, that is minus sine x. Hinglish: Difference quotient me cos(x+h) expand karo aur standard limits lagao: answer -sin x.

Q8. Find the limit of (square root of x minus 1) by (x minus 1) as x tends to 1. Answer: Substitution gives zero over zero. Multiply numerator and denominator by the conjugate (square root of x plus 1); the numerator becomes x minus 1, cancels with the denominator, and the limit is 1 by 2. Hinglish: Conjugate se rationalise karo: (\sqrtx-1)(\sqrtx+1)/(x-1)(\sqrtx+1) = 1/(\sqrtx+1) \to 1/2.

Q9. Differentiate 4x cubed minus 3x plus 2 and evaluate the derivative at x equal to 1. Answer: The derivative is 12x squared minus 3. At x equal to 1 it is 12 minus 3 equal to 9. Differentiating first and substituting afterwards is essential; the derivative of the constant 2 is zero. Hinglish: Derivative 12x² - 3; x = 1 par 9. Pehle differentiate, phir value daalo.

Q10. Find the limit of (x squared minus 1) by (x minus 1) as x tends to 1, and compare it with the function value at x equal to 1. Answer: Factoring gives x plus 1, so the limit is 2. The function is undefined at x equal to 1, because the denominator is zero there, yet the limit exists. This is the standard illustration that a limit describes behaviour near the point, not the value at the point. Hinglish: (x²-1)/(x-1) = x+1 \to limit 2, jabki x = 1 par function undefined hai. Limit point ki value se nahi, behaviour se related hai.

Finally, keep a list of the three standard limits and the six standard derivatives at the top of your rough sheet. Most of the chapter reduces to those nine facts plus algebra. A habit that prevents most errors in this chapter: after computing a limit, substitute a value very close to the point and compare with the answer. If the numerical value is close, the algebra is probably right; if it is far, a factor or a sign is wrong. In limits, always write the substitution step that shows the indeterminate form before doing any algebra. That line makes the strategy visible and prevents accidentally treating zero over zero as a value. In derivatives, write the rule being used above each line, so a reviewer can follow the computation and a dropped term is easy to see. Common mistakes on this chapter Setting h or x to the limiting value before cancelling the common factor turns a finite limit into zero or an undefined expression. Forgetting the negative signs in the derivatives of cosine, cotangent and cosecant is the second error. Mixing limits with values is the third, especially for functions that are undefined at the point. Using degrees instead of radians in a trigonometric limit is the fourth. The first-principles questions need a clean difference quotient line before any limit. Keep the algebra in one column and the limit step below it, so the cancellation is visible. In derivative-at-a-point questions, differentiate first and substitute afterwards. A quick degree check and a sign check catch most remaining errors. Quiz The quiz below has five multiple choice questions on limits, standard trigonometric limits and derivatives. Attempt it after the worked questions above, then move to the previous year page.