Limit questions by short tricks

Limit questions by short tricks

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Limit questions by short tricks

Limit Questions को हल करने के लिए अगर आप लंबे steps से थक चुके हैं, तो यह लेख आपके लिए है। यहाँ हम बताएंगे कि कैसे आप Limit के कठिन से कठिन सवालों को Short Tricks और Smart Methods से कुछ ही सेकंड में हल कर सकते हैं। इस लेख में आप सीखेंगे basic to advanced limit concepts, जैसे — और type questions L’Hospital Rule, factorization और rationalization के short methods , , exponential और trigonometric limits साथ ही आपको मिलेंगे exam-oriented tricks, जो खास तौर पर UP TGT, PGT, LT Grade, SSC, NDA, IIT-JEE, NEET, और Class 11–12 Students के लिए बेहद उपयोगी हैं। इस article के अंत में आप limit के हर प्रकार के सवाल को तेज़ी और accuracy से हल करना सीख जाएंगे। Limits Q1–Q25

Limits — Objective Questions (Q1–Q25)

Math formulas rendered with MathJax. हर सवाल के बाद सही उत्तर और छोटा Trick दिया गया है — उपयोग में सरल।

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Q1.

\( \displaystyle \lim_{x\to0}\frac{e^{1/x}}{e^{1/x+1}} \) is equal to

(a) 0

(b) \( \tfrac{1}{e} \)

(c) Does not exist

(d) None of these

Answer: \( \boxed{\tfrac{1}{e}} \)
Short Trick: Use exponent rules: \( \frac{e^{1/x}}{e^{1/x+1}} = e^{1/x-(1/x+1)}=e^{-1}=\tfrac{1}{e}\).
Q2.

\( \displaystyle \lim_{x\to\pi/6}\frac{\sin 2x}{\sin x} \)

(a) \( \tfrac{1}{\sqrt{3}} \)

(b) \( \tfrac{1}{2} \)

(c) \( \sqrt{3} \)

(d) 1

Answer: \( \boxed{\sqrt{3}} \)
Short Trick: \( \sin2x=2\sin x\cos x \Rightarrow \frac{\sin2x}{\sin x}=2\cos x\). Put \(x=\pi/6\): \(2\cos(\pi/6)=\sqrt3\).
Q3.

\( \displaystyle \lim_{x\to0}\frac{3^{x}-2^{x}}{x} \)

(a) 0

(b) 1

(c) \( \ln\frac{3}{2} \)

(d) \( \tfrac{3}{2} \)

Answer: \( \boxed{\ln\frac{3}{2}} \)
Short Trick: For small \(x\): \(a^x\approx1+x\ln a\). So numerator \(\approx x(\ln3-\ln2)\) → divide by \(x\).
Q4.

\( \displaystyle \lim_{x\to a}\frac{\sqrt{3x-a}-\sqrt{x+a}}{x-a} \)

(a) \( \dfrac{1}{2\sqrt{2a}} \)

(b) \( \dfrac{1}{\sqrt{2a}} \)

(c) \( \dfrac{1}{2\sqrt{a}} \)

(d) \( 2\sqrt{2a} \)

Answer: \( \boxed{\dfrac{1}{2\sqrt{2a}}} \)
Short Trick: Multiply numerator & denominator by conjugate: simplify to \(\dfrac{2x-2a}{(x-a)(\sqrt{3x-a}+\sqrt{x+a})}\) → evaluate at \(x=a\).
Q5.

If \( \displaystyle \lim_{x\to0}(1+3x)^{1/x}=k\), then for continuity at \(x=0\), \(k\) is

(a) -3

(b) 3

(c) \(e^{-3}\)

(d) \(e^{3}\)

Answer: \( \boxed{e^{3}} \)
Short Trick: Standard limit: \(\lim_{t\to0}(1+t)^{1/t}=e\). Put \(t=3x\) → result \(e^3\).
Q6.

\( \displaystyle \lim_{n\to0}\frac{\sqrt{1-n^2}-\sqrt{1+n^2}}{n^2} \)

(a) -2

(b) -1

(c) 0

(d) -1/2

Answer: \( \boxed{-1} \)
Short Trick: Use binomial/series: \( \sqrt{1\pm n^2}\approx1\pm \tfrac{n^2}{2}\). Subtract → \(-n^2\). Divide by \(n^2\) → -1.
Q7.

\( \displaystyle \lim_{x\to1}\frac{2x^2+x-3}{3x^3-3x^2+2x-2} \)

(a) -2

(b) -1

(c) 1

(d) 2

Answer: \( \boxed{1} \)
Short Trick: 0/0 form → differentiate (L’Hôpital) or factor (x-1). Derivatives: numerator’ at 1 =5, denominator’ at 1 =5 → ratio 1.
Q8.

Evaluate \( \displaystyle \lim_{x\to1}\frac{(\sqrt{x}-1)(2x-3)}{2x^2+x-3} \)

(a) -1/10

(b) -1/5

(c) -2/15

(d) 0

Answer: \( \boxed{-\tfrac{1}{10}} \)
Short Trick: Put \(x=1+h\). Expand \(\sqrt{1+h}\approx1+\tfrac h2\). Leading terms give \(-\tfrac{1}{10}\).
Q9.

\( \displaystyle \lim_{x\to2}\frac{x^6-24x-16}{x^3+2x-12} \)

(a) 4

(b) 8

(c) 12

(d) 16

Answer: \( \boxed{12} \)
Short Trick: 0/0 → L’Hôpital: numerator’ at 2 = \(6\cdot2^5-24=168\), denominator’ at 2 = \(3\cdot2^2+2=14\). Ratio =168/14=12.
Q10.

\( \displaystyle \lim_{x\to0}(1+3x)^{1/x} \)

(a) \(e^2\)

(b) \(3e^2\)

(c) \(e^3\)

(d) \(3e^5\)

Answer: \( \boxed{e^{3}} \)
Short Trick: Standard limit: \((1+3x)^{1/x}=(1+3x)^{1/(3x)\cdot3}\to e^{3}.\)
Q11.

\( \displaystyle \lim_{x\to0}\frac{\ln(1+x)}{3^x-1} \)

(a) 1

(b) \( \log_2 3\) (scan)

(c) \( \dfrac{1}{\ln 3} \)

(d) \( \log_3 e\)

Answer: \( \boxed{\dfrac{1}{\ln 3}} \)
Short Trick: For small \(x\): \(\ln(1+x)\sim x,\;3^x-1\sim x\ln3\). Ratio = \(1/\ln3\).
Q12.

\( \displaystyle \lim_{x\to2}\frac{\sqrt{x-2}+\sqrt{x}-\sqrt{2}}{\sqrt{x^2-4}} \)

(a) 2

(b) 1

(c) 1/2

(d) 0

Answer: \( \boxed{\tfrac{1}{2}} \)
Short Trick: Put \(x=2+h\). Leading order: numerator \(\sim\sqrt h\), denominator \(\sim2\sqrt h\). Ratio \(\to\frac12\).
Q13.

\( \displaystyle \lim_{x\to0}\frac{8^x-4^x-2^x+1}{x^2} \)

(a) 1

(b) \(2\ln2\)

(c) \(2(\ln2)^2\)

(d) \(3(\ln2)^3\)

Answer: \( \boxed{2(\ln2)^2} \)
Short Trick: Expand each \(a^x\) to second order: \(1+x\ln a+\tfrac{x^2}{2}(\ln a)^2\). First-order cancels; compute second-order term.
Q14.

\( \displaystyle \lim_{n\to\infty}\frac{1^2+2^2+\cdots+n^2}{n^3} \)

(a) 0

(b) 1/3

(c) 3

(d) None of these

Answer: \( \boxed{\tfrac{1}{3}} \)
Short Trick: Sum = \(\frac{n(n+1)(2n+1)}{6}\sim \frac{n^3}{3}\). Divide by \(n^3\) → \(1/3\).
Q15.

\( \displaystyle \lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^3} \)

(a) 0

(b) 1

(c) 1/3

(d) 2/3

Answer: \( \boxed{\tfrac{1}{3}} \)
Short Trick: Same as Q14 (use formula for sum of squares).
Q16.

\( \displaystyle \lim_{x\to0}\frac{a^x-b^x}{e^x-1} \) (a,b>0)

(a) \( \ln\frac{a}{b} \)

(b) \( \ln\frac{b}{a} \)

(c) \( \ln(ab) \)

(d) \( \ln(a+b) \)

Answer: \( \boxed{\ln\frac{a}{b}} \)
Short Trick: For small \(x\): \(a^x-b^x\approx x(\ln a-\ln b)\) and \(e^x-1\approx x\). Ratio = \(\ln a-\ln b\).
Q17.

\( \displaystyle \lim_{\theta\to0}\frac{\cos5\theta-\cos7\theta}{\theta^2} \)

(a) -12

(b) -6

(c) 6

(d) 12

Answer: \( \boxed{12} \)
Short Trick: Use identity: \(\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\). For small \(\theta\) reduce → \(12\theta^2\).
Q18.

\( \displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2} \)

(a) 0

(b) 1/4

(c) 1/2

(d) 1

Answer: \( \boxed{\tfrac{1}{2}} \)
Short Trick: Use expansion: \(1-\cos x\approx \tfrac{x^2}{2}\).
Q19.

\( \displaystyle \lim_{x\to2}\frac{3^{x/2}-3}{3^{x}-9} \)

(a) \(3^{-2}\)

(b) 1/3

(c) 1/6

(d) \( \ln 3\)

Answer: \( \boxed{\tfrac{1}{6}} \)
Short Trick: 0/0 → derivative ratio: numerator’ = \(\tfrac{\ln3}{2}3^{x/2}\), denom’ = \(\ln3\cdot3^x\). At \(x=2\) gives \(\tfrac{(\ln3)3/2}{(\ln3)9}=\tfrac{1}{6}\).
Q20.

If \( \displaystyle \lim_{x\to5}\frac{x^k-5^k}{x-5}=500\), then \(k\) equals

(a) 3

(b) 4

(c) 5

(d) 6

Answer: \( \boxed{4} \)
Short Trick: Limit = derivative at 5 = \(k\cdot5^{k-1}=500\). Try \(k=4\): \(4\cdot125=500\).
Q21.

\( \displaystyle \lim_{x\to3\pi}\frac{1+\tan x}{\cos 2x} \)

(a) -2

(b) -1

(c) 0

(d) 1

Answer: \( \boxed{1} \)
Short Trick: Use periodicity: at \(x=3\pi\): \(\tan(3\pi)=0,\;\cos(6\pi)=1\). So \((1+0)/1=1\).
Q22.

\( \displaystyle \lim_{x\to0}\frac{5^x-5^{-x}}{2x} \)

(a) 0

(b) 1

(c) \(\ln 5\)

(d) \(2\ln5\)

Answer: \( \boxed{\ln 5} \)
Short Trick: Expand: \(5^x\approx1+x\ln5\), \(5^{-x}\approx1-x\ln5\). Numerator \(\approx2x\ln5\). Divide by \(2x\).
Q23.

If \( f(x)=\begin{cases}1+x,&x>0\\ x,&x<0\end{cases}\) then \(\lim_{x\to0}f(x)=\ ?\)

(a) 0

(b) 1/2

(c) 1

(d) Non-existent

Answer: \( \boxed{\text{Non-existent}} \)
Short Trick: Right-hand limit =1, left-hand =0 → limits differ → does not exist.
Q24.

If \( f(x)=\begin{cases} x,&x<0\\ 1,&x=0\\ x^2,&x>0 \end{cases}\) then \(\lim_{x\to0} f(x)=\ ?\)

(a) 0

(b) 1

(c) 2

(d) Does not exist

Answer: \( \boxed{0} \)
Short Trick: Left limit =0, right limit =0 → overall limit =0 (value at 0 is 1 but limit is 0).
Q25.

\( \displaystyle \lim_{x\to\infty} x\big(\sqrt{x^2+6}-x\big) \)

(a) -6

(b) -3

(c) 3

(d) 4

Answer: \( \boxed{3} \)
Short Trick: Multiply by conjugate: \(x(\sqrt{x^2+6}-x) = x\cdot \dfrac{6}{\sqrt{x^2+6}+x} \sim x\cdot \dfrac{6}{2x}=3.\)

End of Part 1 (Q1–Q25). → बताइए क्या मैं अब Part 2 (Q26–Q50) भेज दूँ?


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