Limits Important Questions for JEE Mains, Advanced, AIEEE
Q1. $\lim_{x\rightarrow 0}\frac {log_{e}(1+3x)-log_{e}(3-x)}{x}=?$[AIEEE 2003]
Ans:
Shortcut method:
$ \lim_{x\rightarrow 0}\frac {log_{e}(1+ax)-log_{e}(a-x)}{x}=\frac{2}{a}$
Put $a=3$
$\lim_{x\rightarrow 0}\frac {log_{e}(1+3x)-log_{e}(3-x)}{x}=\frac{2}{3}$
Q2.$ \lim_{x\rightarrow \infty} (1+\frac{1}{x})^{\frac{1}{x}} = ?$ [JEE 2007]
Ans: Shortcut method:
Applying Formula 10
$ =\lim_{x\rightarrow \infty} e^{(1+\frac{1}{x}-1)\frac{1}{x}}$
$ =e^{\frac{1}{2}}$
Q3. $ \lim_{x\rightarrow \infty} (\frac{x+6}{x+1})^{x+4} =? $ [JEE 2004, IIT 1990]
$= \lim_{x\rightarrow \infty} (1+\frac{5}{x+1})^{x+1+3}$
Now,
Case:1
=$ \lim_{x+1\rightarrow \infty} (1+\frac{5}{x+1})^{x+1}$
=$ e^5$
Case:2
$ =\lim_{x\rightarrow \infty} (1+\frac{5}{x+1})^{3}$
$ = 1 $
So, $\small \lim_{x\rightarrow \infty} (\frac{x+6}{x+1})^{x+4}=3$
Q4.$ \lim_{x\rightarrow 0} (\frac{2^x-1}{\sqrt {x+1} -1})$=?[JEE 2010, IIT 1982]
Ans:
Dividing by $\small x$ for this limit $\small \lim_{x\rightarrow 0} (\frac{2^x-1}{\sqrt {x+1} -1})$
Trick:
Case1:
$ \lim_{x\rightarrow 0} \frac{2^x-1}{x}=log_{e}2$
Case2:
$\small \lim_{x\rightarrow 0} \frac{\sqrt {x+1} -1}{x+1-1} = \frac{1}{2}$
Ans: $\small log_{e}2$
Q5.$\small \lim_{x\rightarrow 0} \frac{xe^x-log(1+x)}{x^2}=? $[WBJEE]
Ans:
$= \lim_{x\rightarrow 0} \frac{xe^x-x+x-log(1+x)}{x^2}$
$\small \lim_{x\rightarrow 0} [\frac{xe^x-x}{x^2}+\frac{x-(x-\frac{x^2}{2}+\frac{x^3}{3}-..)}{x^2}]$
$= 1+\small \frac{1}{2}+0+0+....$
Q6.$\small \lim_{x\rightarrow \infty} (\frac{5x^2+1}{3x^2+1})^{\frac{1}{x^2}}$=? [IIT 1996, WBJEE 2010]
Ans:
=$\small e^{\lim_{x\rightarrow \infty} (\frac{5x^2+1}{3x^2+1}-1)^{\frac{1}{x^2}}}$
=$\small e^{\lim_{x\rightarrow \infty} (\frac{2}{3x^2+1})^{\frac{1}{x^2}}}$
=$\small e^2$
Q7.$\small \lim_{n\rightarrow \infty} (\frac{5^n+3^n}{5^n-3^n})$=?
a. 1
b. 2
c. $\small \frac{1}{2}$
d. $\small \frac{1}{3}$
Ans: (a)
Hint:
$\small \lim_{n\rightarrow \infty} (\frac{5^n+3^n}{5^n-3^n})$
=$\small \lim_{n\rightarrow \infty} \frac{1+(\frac{3}{5})^n}{1-(\frac{3}{5})^n}$
Since $\small \frac{3}{5}<1$, and $\small n\rightarrow \infty$, and $(\frac{3}{5})\rightarrow 0$
Q8. $\small \lim_{n\rightarrow \infty} [cos\frac{x}{n}]^n$=?
a. -1
b. 0
c. 1
d. None of these
Ans: 1
Hint:
$\small \lim_{n\rightarrow \infty} [cos\frac{x}{n}]^n$
$\small \lim_{n\rightarrow \infty} [1-2 sin^2\frac{x}{2n}]^n$
Apply Formula 10, when $\small f(n)=\small 1-2 sin^2\frac{x}{2n},g(n)=n$
Find, $\small \lim_{n\rightarrow \infty}f(n)^{g(n)}$
Q9. $\small \lim_{x\rightarrow 0^+}\frac{e^{\left [ x \right ]+\left | x \right |}-1}{\left [ x \right ]+\left | x \right |}$=?
a. 0
b. -1
c. 1
Ans: 1
Q10.$\small \lim_{x\rightarrow 0} (\frac{sinx}{x})^{\frac{sinx}{x-sinx}}$=?
a. $\small e$
b.$\small e^{-1}$
c.$\small e^2$
d.$\small e^{-2}$
Ans: $\small e^{-1}$
Hint: Apply formula 10
Q11.$\small \lim_{x\rightarrow 0} [tan(\frac{\pi}{4}+x)]^{\frac{1}{x}}$=?
a.$\small e$
b.$\small e^3$
c.$\small e^2$
d.$\small e^{-1}$
Ans: $\small e^2$
Hint:
Formula 10
Q12.$\small \frac{sin|x|}{|x|}$=?
a.1
b.-1
c.0
d. Does not exist
Ans: (d)
Q13.$\small \lim_{x\rightarrow 0} \frac{e^{x^2}-cosx}{x^2}=?$
a. $\small \frac{1}{2}$
b. 1
c. $\small \frac{2}{3}$
d.$\small \frac{3}{2}$
Ans: (d)
Hint: L'Hospitals Rule $\small [\frac{0}{0}]$ form
Q145.$\small\lim_{x\rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt 2 -\sqrt(1+cosx)}$=?
a. $\small 4\sqrt 2(log 3)^2$
b. $\small 8\sqrt 2(log 3)^2$
c. $\small 12\sqrt 2(log 3)^2$
d. $\small 2\sqrt 2(log 3)^2$
Ans:(b)
Q15.$\small \lim_{x\rightarrow 1}\frac{sin(e^{x-1}-1)}{log_{e}x} $=?
a. 0
b. $\small e$
c. $\small \frac {1}{e}$
d.$\small 1$
Ans: d
Hint: Let, $\small z=x-1$
$ =\lim_{x\rightarrow \infty} e^{(1+\frac{1}{x}-1)\frac{1}{x}}$
$ =e^{\frac{1}{2}}$
Q3. $ \lim_{x\rightarrow \infty} (\frac{x+6}{x+1})^{x+4} =? $ [JEE 2004, IIT 1990]
$= \lim_{x\rightarrow \infty} (1+\frac{5}{x+1})^{x+1+3}$
Now,
Case:1
=$ \lim_{x+1\rightarrow \infty} (1+\frac{5}{x+1})^{x+1}$
=$ e^5$
Case:2
$ =\lim_{x\rightarrow \infty} (1+\frac{5}{x+1})^{3}$
$ = 1 $
So, $\small \lim_{x\rightarrow \infty} (\frac{x+6}{x+1})^{x+4}=3$
Q4.$ \lim_{x\rightarrow 0} (\frac{2^x-1}{\sqrt {x+1} -1})$=?[JEE 2010, IIT 1982]
Ans:
Dividing by $\small x$ for this limit $\small \lim_{x\rightarrow 0} (\frac{2^x-1}{\sqrt {x+1} -1})$
Trick:
Case1:
$ \lim_{x\rightarrow 0} \frac{2^x-1}{x}=log_{e}2$
Case2:
$\small \lim_{x\rightarrow 0} \frac{\sqrt {x+1} -1}{x+1-1} = \frac{1}{2}$
Ans: $\small log_{e}2$
Q5.$\small \lim_{x\rightarrow 0} \frac{xe^x-log(1+x)}{x^2}=? $[WBJEE]
Ans:
$= \lim_{x\rightarrow 0} \frac{xe^x-x+x-log(1+x)}{x^2}$
$\small \lim_{x\rightarrow 0} [\frac{xe^x-x}{x^2}+\frac{x-(x-\frac{x^2}{2}+\frac{x^3}{3}-..)}{x^2}]$
$= 1+\small \frac{1}{2}+0+0+....$
Q6.$\small \lim_{x\rightarrow \infty} (\frac{5x^2+1}{3x^2+1})^{\frac{1}{x^2}}$=? [IIT 1996, WBJEE 2010]
Ans:
=$\small e^{\lim_{x\rightarrow \infty} (\frac{5x^2+1}{3x^2+1}-1)^{\frac{1}{x^2}}}$
=$\small e^{\lim_{x\rightarrow \infty} (\frac{2}{3x^2+1})^{\frac{1}{x^2}}}$
=$\small e^2$
Q7.$\small \lim_{n\rightarrow \infty} (\frac{5^n+3^n}{5^n-3^n})$=?
a. 1
b. 2
c. $\small \frac{1}{2}$
d. $\small \frac{1}{3}$
Ans: (a)
Hint:
$\small \lim_{n\rightarrow \infty} (\frac{5^n+3^n}{5^n-3^n})$
=$\small \lim_{n\rightarrow \infty} \frac{1+(\frac{3}{5})^n}{1-(\frac{3}{5})^n}$
Since $\small \frac{3}{5}<1$, and $\small n\rightarrow \infty$, and $(\frac{3}{5})\rightarrow 0$
Q8. $\small \lim_{n\rightarrow \infty} [cos\frac{x}{n}]^n$=?
a. -1
b. 0
c. 1
d. None of these
Ans: 1
Hint:
$\small \lim_{n\rightarrow \infty} [cos\frac{x}{n}]^n$
$\small \lim_{n\rightarrow \infty} [1-2 sin^2\frac{x}{2n}]^n$
Apply Formula 10, when $\small f(n)=\small 1-2 sin^2\frac{x}{2n},g(n)=n$
Find, $\small \lim_{n\rightarrow \infty}f(n)^{g(n)}$
Q9. $\small \lim_{x\rightarrow 0^+}\frac{e^{\left [ x \right ]+\left | x \right |}-1}{\left [ x \right ]+\left | x \right |}$=?
a. 0
b. -1
c. 1
Ans: 1
Q10.$\small \lim_{x\rightarrow 0} (\frac{sinx}{x})^{\frac{sinx}{x-sinx}}$=?
a. $\small e$
b.$\small e^{-1}$
c.$\small e^2$
d.$\small e^{-2}$
Ans: $\small e^{-1}$
Hint: Apply formula 10
Q11.$\small \lim_{x\rightarrow 0} [tan(\frac{\pi}{4}+x)]^{\frac{1}{x}}$=?
a.$\small e$
b.$\small e^3$
c.$\small e^2$
d.$\small e^{-1}$
Ans: $\small e^2$
Hint:
Formula 10
Q12.$\small \frac{sin|x|}{|x|}$=?
a.1
b.-1
c.0
d. Does not exist
Ans: (d)
Q13.$\small \lim_{x\rightarrow 0} \frac{e^{x^2}-cosx}{x^2}=?$
a. $\small \frac{1}{2}$
b. 1
c. $\small \frac{2}{3}$
d.$\small \frac{3}{2}$
Ans: (d)
Hint: L'Hospitals Rule $\small [\frac{0}{0}]$ form
Q145.$\small\lim_{x\rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt 2 -\sqrt(1+cosx)}$=?
a. $\small 4\sqrt 2(log 3)^2$
b. $\small 8\sqrt 2(log 3)^2$
c. $\small 12\sqrt 2(log 3)^2$
d. $\small 2\sqrt 2(log 3)^2$
Ans:(b)
Q15.$\small \lim_{x\rightarrow 1}\frac{sin(e^{x-1}-1)}{log_{e}x} $=?
a. 0
b. $\small e$
c. $\small \frac {1}{e}$
d.$\small 1$
Ans: d
Hint: Let, $\small z=x-1$
Tags: JEE Mains 2021, WBJEE 2021, AIEEE, Limits Imortant Questions with Answers, Limits Previous Years Solutions