Q: \$\displaystyle \lim_{n \rightarrow \infty}\dfrac {1^{2}+2^{2}+3^{2}+....+n^{2}}{n^{3}}\$ is equal to

Solution\$ \displaystyle \lim_{x\rightarrow \infty } \dfrac{1^2 + 2^2 +--- n^2}{n^3}\$we know that \$ 1^2 + 2^2 + --- n^2 = \dfrac{n (n+1)(2n+1)}{6}\$\$ \displaystyle =\lim_{x\rightarrow \infty }\frac{n(n+1)(2n+1)}{6n^3}\$\$ \displaystyle= \lim_{x\rightarrow \infty }\frac{(n+1)(2n+1)}{6n^2} = \lim_{x\rightarrow \infty } \frac{2n^2 +3n +1}{6n^2}\$Apply L-Hospital rule twice.\$ \displaystyle = \lim_{x\rightarrow \infty } \dfrac{4n+3}{12n}\$\$ = \dfrac{4}{12} = \dfrac{1}{3}\$C is correct.

Q: \$\displaystyle \lim_{x\rightarrow 1}\frac{\left ( 2x-3 \right )\left ( \sqrt{x}-1 \right )}{2x^{2}+x-3}.\$ is \$\displaystyle \frac{-1}{b}\$ Find b

\$\displaystyle \lim_{x\rightarrow 1}\frac{\left ( 2x-3 \right )\left ( \sqrt{x}-1 \right )}{2x^{2}+x-3}\$\$\displaystyle= \lim_{x\rightarrow 1}\frac{\left ( 2x-3 \right )\left ( \sqrt{x}-1 \right )}{\left ( 2x+3 \right )\left ( x-1 \right )}\$\$\displaystyle = \lim_{x\rightarrow 1}\frac{2x-3}{\left ( 2x+3 \right )\left ( \sqrt{x}+1 \right )}\$\$\displaystyle= \frac{2-3}{\left ( 2+3 \right )\left ( 1+1 \right )}\$\$= -\dfrac{1}{10}\$\$\Rightarrow b=10\$

Q: If \$\displaystyle\lim_{x \rightarrow 1}\dfrac {x^{4}-1}{x-1}=\lim_{x \rightarrow k}\dfrac {x^{3}-k^{3}}{ x^{2}-k^{2}}\$, then find the value of \$k\$.

\$\displaystyle \rightarrow \lim_{x\rightarrow 1} \frac{x^{4}-1}{x-1} = \lim_{x\rightarrow 1} \frac{(x^{2}-1)(x^{2}+1)}{x-1} \$\$\displaystyle = \lim_{x\rightarrow 1} \frac{(x-1)(x+1)(x^{2}+1)}{(x-1)} \$\$\displaystyle = (1+1)(1+1) \$\$ \displaystyle= (2) (2) \$\$ \displaystyle= 4 \$ ...(1)Now \$\displaystyle \lim_{x\rightarrow 1} \frac{x^{3}-k^{3}}{x^{2}-k^{2}} \$\$\displaystyle = \lim_{x\rightarrow 1} \frac{(x-k)(x^{2}+xk+k^{2})}{(x-k)(x+k)} \$\$\displaystyle = x^{2}+\frac{k^{2}+k^{2}+k^{2}}{2k} \$\$ \displaystyle= \frac{3k^{2}}{2k} = \frac{3k}{2} \$...(2)\$ \displaystyle \therefore 4 = \frac{3k}{2} \$\$ \displaystyle \therefore k = \frac{8}{3} \$

Q: Solve \$\lim _ { x \rightarrow 0 } \dfrac { \sqrt { 2 + x } - \sqrt { 2 } } { x }\$

\$\\\lim_{x\rightarrow0}(\frac{\sqrt{2+x}-\sqrt{2}}{x})\times(\frac{\sqrt{2+x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{2}})\\=\lim_{x\rightarrow0}(\frac{(2+x)-2}{x\left(\sqrt{2+x}+\sqrt{2}\right)})\\=\lim_{x\rightarrow0}(\frac{1}{\left(\sqrt{2+x}+\sqrt{2}\right)})\\=(\frac{1}{2\sqrt{2}})\$

Q: Evaluate \$\displaystyle \lim_{x \rightarrow 1}\left( \dfrac{1}{x^2 + x - 2} - \dfrac{x}{x^3 - 1}\right)\$

We need to find value of \$\lim _{ x\rightarrow 1 }{ \left( \dfrac { 1 }{ { x }^{ 2 }+x-2 } -\dfrac { x }{ { x }^{ 3 }-1 } \right) } \$\$= \lim _{ x\rightarrow 1 }{ \left( \dfrac { 1 }{ (x-1)(x+2) } -\dfrac { x }{ { (x-1)(x }^{ 2 }+1+x) } \right) } \\ =\lim _{ x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { 1 }{ x+2 } -\dfrac { x }{ { x }^{ 2 }+x+1 } \right) \right\} } \\= \lim _{ x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { { x }^{ 2 }+x+1-{ x }^{ 2 }-2x }{ (x+2)({ x }^{ 2 }+x+1) } \right) \right\} } \\= \lim _{ x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { 1-x }{ (x+2)({ x }^{ 2 }+x+1) } \right) \right\} } \\= \lim _{ x\rightarrow 1 }{ \dfrac { -1 }{ (x+2)({ x }^{ 2 }+x+1) } } \\ =\dfrac { -1 }{ (1+2)(1+1+1) } \\ =-\dfrac { 1 }{ 9 } \$

Q: Evaluate: \$\mathop {\lim }\limits_{x \to 2 } \left[ {\dfrac{{{x^3} - 4{x^2} + 4x}}{{{x^2} - 4}}} \right]\$

\$\displaystyle \lim_{x \rightarrow 2} \dfrac{x^{3}-4x^{2}+4x}{x^{2}-4}= \lim_{x \rightarrow 2} \dfrac{x(x^{2}-4x+4)}{(x-2)(x+2)}\$\$\displaystyle=\lim_{x \rightarrow 2} \dfrac{x(x-2)^{2}}{(x-2)(x+2)}\$\$\displaystyle=\lim_{x \rightarrow 2} \dfrac{x(x-2)}{(x+2)}=0\$

Q: The value of \$\displaystyle \lim_{x\rightarrow 2} \dfrac {(x^{4} - 16)}{x - 2} \$ is

The value of the limit \$\displaystyle \lim _{ x\rightarrow 2 }{ \dfrac { { x }^{ 4 }-16 }{ x-2 } } \$ is\$=\displaystyle \lim _{ x\rightarrow 2 }{ \dfrac { ({ x }^{ 2 }-4)({ x }^{ 2 }+4) }{ x-2 } } \$\$=\displaystyle \lim _{ x\rightarrow 2 }{ \dfrac { ({ x }-2)(x+2)({ x }^{ 2 }+4) }{ x-2 } } \$\$=\displaystyle \lim _{ x\rightarrow 2 }{ (x+2)({ x }^{ 2 }+4) } \$\$=32\$

Q: If \$\displaystyle\lim_{x\rightarrow 3}\dfrac{x^n-3^n}{x-3}=108\$, find the value of n.

Using formula \$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^n-a^n}{x-a}}=n.a^{n-1}\$\$\displaystyle \lim_{x\rightarrow 3}{\dfrac{x^n-3^n}{x-3}}=n.3^{4-1}=4.3^3\$\$n=4\$

Q: Solve the equation:-\$\lim_{{x\rightarrow \pi/2}}\ \ \dfrac{\tan\ 2x}{x-\pi/2}\$

\${\lim}_{x\rightarrow\,\frac{\pi}{2}}{\dfrac{\tan{2x}}{x-\dfrac{\pi}{2}}}\$\$LHL={\lim}_{h\rightarrow\,{0}^{-}}{\dfrac{\tan{2\left(\dfrac{\pi}{2}-h\right)}}{\dfrac{\pi}{2}-h-\dfrac{\pi}{2}}}\$\$={\lim}_{h\rightarrow\,{0}^{-}}{\dfrac{\tan{\left(\pi-2h\right)}}{-h}}\$\$={\lim}_{h\rightarrow\,{0}^{-}}{-\dfrac{\tan{2h}}{-h}}\$\$={\lim}_{h\rightarrow\,{0}^{-}}{-\dfrac{2\tan{2h}}{-2h}}=2\$\$RHL={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{2\left(\dfrac{\pi}{2}+h\right)}}{\dfrac{\pi}{2}+h-\dfrac{\pi}{2}}}\$\$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{\left(\pi+2h\right)}}{h}}\$\$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{2h}}{h}}\$\$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{2\tan{2h}}{2h}}=2\$Thus,\${\lim}_{x\rightarrow\,\frac{\pi}{2}}{\dfrac{\tan{2x}}{x-\dfrac{\pi}{2}}}=2\$

Question 1

$\displaystyle \lim_{n \rightarrow \infty}\dfrac {1^{2}+2^{2}+3^{2}+....+n^{2}}{n^{3}}$ is equal to

Accepted Answer

Solution$ \displaystyle \lim_{x\rightarrow \infty } \dfrac{1^2 + 2^2 +--- n^2}{n^3}$we know that $ 1^2 + 2^2 + --- n^2 = \dfrac{n (n+1)(2n+1)}{6}$$ \displaystyle =\lim_{x\rightarrow \infty }\frac{n(n+1)(2n+1)}{6n^3}$$ \displaystyle= \lim_{x\rightarrow \infty }\frac{(n+1)(2n+1)}{6n^2} = \lim_{x\rightarrow \infty } \frac{2n^2 +3n +1}{6n^2}$Apply L-Hospital rule twice.$ \displaystyle = \lim_{x\rightarrow \infty } \dfrac{4n+3}{12n}$$ = \dfrac{4}{12} = \dfrac{1}{3}$C is correct.

Question 2

$\displaystyle \lim_{x\rightarrow 1}\frac{\left ( 2x-3 \right )\left ( \sqrt{x}-1 \right )}{2x^{2}+x-3}.$ is $\displaystyle \frac{-1}{b}$ Find b

Accepted Answer

$\displaystyle \lim_{x\rightarrow 1}\frac{\left ( 2x-3 \right )\left ( \sqrt{x}-1 \right )}{2x^{2}+x-3}$$\displaystyle= \lim_{x\rightarrow 1}\frac{\left ( 2x-3 \right )\left ( \sqrt{x}-1 \right )}{\left ( 2x+3 \right )\left ( x-1 \right )}$$\displaystyle = \lim_{x\rightarrow 1}\frac{2x-3}{\left ( 2x+3 \right )\left ( \sqrt{x}+1 \right )}$$\displaystyle= \frac{2-3}{\left ( 2+3 \right )\left ( 1+1 \right )}$$= -\dfrac{1}{10}$$\Rightarrow b=10$

Question 3

If $\displaystyle\lim_{x  \rightarrow 1}\dfrac {x^{4}-1}{x-1}=\lim_{x \rightarrow k}\dfrac {x^{3}-k^{3}}{ x^{2}-k^{2}}$,  then find the value of $k$.

Accepted Answer

$\displaystyle \rightarrow \lim_{x\rightarrow 1} \frac{x^{4}-1}{x-1} = \lim_{x\rightarrow 1} \frac{(x^{2}-1)(x^{2}+1)}{x-1} $$\displaystyle = \lim_{x\rightarrow 1} \frac{(x-1)(x+1)(x^{2}+1)}{(x-1)} $$\displaystyle = (1+1)(1+1) $$ \displaystyle= (2) (2) $$ \displaystyle= 4 $ ...(1)Now $\displaystyle \lim_{x\rightarrow 1} \frac{x^{3}-k^{3}}{x^{2}-k^{2}} $$\displaystyle = \lim_{x\rightarrow 1} \frac{(x-k)(x^{2}+xk+k^{2})}{(x-k)(x+k)} $$\displaystyle = x^{2}+\frac{k^{2}+k^{2}+k^{2}}{2k} $$ \displaystyle= \frac{3k^{2}}{2k} = \frac{3k}{2} $...(2)$ \displaystyle \therefore 4 = \frac{3k}{2} $$ \displaystyle \therefore k = \frac{8}{3} $

Question 4

Solve $\lim _ { x \rightarrow 0 } \dfrac { \sqrt { 2 + x } - \sqrt { 2 } } { x }$

Accepted Answer

$\\lim_{x\rightarrow0}(\frac{\sqrt{2+x}-\sqrt{2}}{x})\times(\frac{\sqrt{2+x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{2}})\=\lim_{x\rightarrow0}(\frac{(2+x)-2}{x\left(\sqrt{2+x}+\sqrt{2}\right)})\=\lim_{x\rightarrow0}(\frac{1}{\left(\sqrt{2+x}+\sqrt{2}\right)})\=(\frac{1}{2\sqrt{2}})$

Question 5

Evaluate $\displaystyle \lim_{x \rightarrow 1}\left( \dfrac{1}{x^2 + x - 2} - \dfrac{x}{x^3 - 1}\right)$

Accepted Answer

We need to find value of $\lim _{ x\rightarrow 1&#10;}{ \left( \dfrac { 1 }{ { x }^{ 2 }+x-2 } -\dfrac { x }{ { x }^{ 3 }-1 } &#10;\right)  } $$= \lim _{ x\rightarrow 1 }{ \left( \dfrac { 1 }{ (x-1)(x+2) }&#10;-\dfrac { x }{ { (x-1)(x }^{ 2 }+1+x) }  \right)  } \ =\lim _{&#10;x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { 1 }{ x+2 } -\dfrac&#10;{ x }{ { x }^{ 2 }+x+1 }  \right)  \right\}  } \= \lim _{&#10;x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { { x }^{ 2 }+x+1-{&#10;x }^{ 2 }-2x }{ (x+2)({ x }^{ 2 }+x+1) }  \right)  \right\}  }&#10;\= \lim _{ x\rightarrow 1 }{ \left\{ \dfrac { 1 }{ (x-1) } \left( \dfrac { 1-x&#10;}{ (x+2)({ x }^{ 2 }+x+1) }  \right)  \right\}  } \= \lim _{&#10;x\rightarrow 1 }{ \dfrac { -1 }{ (x+2)({ x }^{ 2 }+x+1) }  } \ =\dfrac {&#10;-1 }{ (1+2)(1+1+1) } \ =-\dfrac { 1 }{ 9 } $

Question 6

Evaluate: $\mathop {\lim }\limits_{x \to 2 } \left[ {\dfrac{{{x^3} - 4{x^2} + 4x}}{{{x^2} - 4}}} \right]$

Accepted Answer

$\displaystyle \lim_{x \rightarrow 2} \dfrac{x^{3}-4x^{2}+4x}{x^{2}-4}= \lim_{x \rightarrow 2} \dfrac{x(x^{2}-4x+4)}{(x-2)(x+2)}$$\displaystyle=\lim_{x \rightarrow 2} \dfrac{x(x-2)^{2}}{(x-2)(x+2)}$$\displaystyle=\lim_{x \rightarrow 2} \dfrac{x(x-2)}{(x+2)}=0$

Question 7

If $f(x)=x^{3}-\dfrac{1}{x^{3}}$, show that $f(x)+f\left(\dfrac{1}{x}\right)=0$

Accepted Answer

Toppr Ask Answer

Question 8

The value of $\displaystyle \lim_{x\rightarrow 2} \dfrac {(x^{4} - 16)}{x - 2} $ is

Accepted Answer

The value of the limit $\displaystyle \lim _{ x\rightarrow 2 }{ \dfrac { { x }^{ 4 }-16 }{ x-2 }  } $ is$=\displaystyle \lim _{ x\rightarrow 2 }{ \dfrac { ({ x }^{ 2 }-4)({ x }^{ 2 }+4) }{ x-2 }  } $$=\displaystyle \lim _{ x\rightarrow 2 }{ \dfrac { ({ x }-2)(x+2)({ x }^{ 2 }+4) }{ x-2 }  } $$=\displaystyle \lim _{ x\rightarrow 2 }{ (x+2)({ x }^{ 2 }+4) } $$=32$

Question 9

If $\displaystyle\lim_{x\rightarrow 3}\dfrac{x^n-3^n}{x-3}=108$, find the value of n.

Accepted Answer

Using formula $\displaystyle \lim_{x\rightarrow a}{\dfrac{x^n-a^n}{x-a}}=n.a^{n-1}$$\displaystyle \lim_{x\rightarrow 3}{\dfrac{x^n-3^n}{x-3}}=n.3^{4-1}=4.3^3$$n=4$

Question 10

Solve the equation:-$\lim_{{x\rightarrow \pi/2}}\ \ \dfrac{\tan\ 2x}{x-\pi/2}$

Accepted Answer

${\lim}_{x\rightarrow\,\frac{\pi}{2}}{\dfrac{\tan{2x}}{x-\dfrac{\pi}{2}}}$$LHL={\lim}_{h\rightarrow\,{0}^{-}}{\dfrac{\tan{2\left(\dfrac{\pi}{2}-h\right)}}{\dfrac{\pi}{2}-h-\dfrac{\pi}{2}}}$$={\lim}_{h\rightarrow\,{0}^{-}}{\dfrac{\tan{\left(\pi-2h\right)}}{-h}}$$={\lim}_{h\rightarrow\,{0}^{-}}{-\dfrac{\tan{2h}}{-h}}$$={\lim}_{h\rightarrow\,{0}^{-}}{-\dfrac{2\tan{2h}}{-2h}}=2$$RHL={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{2\left(\dfrac{\pi}{2}+h\right)}}{\dfrac{\pi}{2}+h-\dfrac{\pi}{2}}}$$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{\left(\pi+2h\right)}}{h}}$$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{\tan{2h}}{h}}$$={\lim}_{h\rightarrow\,{0}^{+}}{\dfrac{2\tan{2h}}{2h}}=2$Thus,${\lim}_{x\rightarrow\,\frac{\pi}{2}}{\dfrac{\tan{2x}}{x-\dfrac{\pi}{2}}}=2$

Basic Numericals on Algebra of Limits.

Language

Rate

REVISE WITH CONCEPTS

Algebra of Limits

Algebra of Limits - Proofs

QUICK SUMMARY WITH STORIES

Limits by Direct Substitution

Algebra of limits of polynomial and rational function

$n \to \infty lim \frac{1 ^{2} + 2 ^{2} + 3 ^{2} + . . . . + n ^{2}}{n ^{3}}$ is equal to

$x \to 1 lim \frac{( 2 x - 3 ) ( x - 1 )}{2 x ^{2} + x - 3} .$ is $\frac{- 1}{b}$ Find b

If $x \to 1 lim \frac{x ^{4} - 1}{x - 1} = x \to k lim \frac{x ^{3} - k ^{3}}{x ^{2} - k ^{2}}$ , then find the value of $k$ .

Solve $lim_{x \to 0} \frac{2 + x - 2}{x}$

Evaluate $x \to 1 lim (\frac{1}{x ^{2} + x - 2} - \frac{x}{x ^{3} - 1})$

Evaluate: $x \to 2 lim [\frac{x ^{3} - 4 x ^{2} + 4 x}{x ^{2} - 4}]$

If $f (x) = x^{3} - \frac{1}{x ^{3}}$ , show that $f (x) + f (\frac{1}{x}) = 0$

The value of $x \to 2 lim \frac{( x ^{4} - 1 6 )}{x - 2}$ is

If $x \to 3 lim \frac{x ^{n} - 3 ^{n}}{x - 3} = 108$ , find the value of n.

Solve the equation:-
$lim_{x \to π / 2} \frac{tan 2 x}{x - π / 2}$

Basic Numericals on Algebra of Limits.

Language

Rate

REVISE WITH CONCEPTS

Algebra of Limits

Algebra of Limits - Proofs

QUICK SUMMARY WITH STORIES

Limits by Direct Substitution

Algebra of limits of polynomial and rational function

n→∞lim​n312+22+32+....+n2​ is equal to

x→1lim​2x2+x−3(2x−3)(x​−1)​. is b−1​ Find b

If x→1lim​x−1x4−1​=x→klim​x2−k2x3−k3​, then find the value of k.

Solve limx→0​x2+x​−2​​

Evaluate x→1lim​(x2+x−21​−x3−1x​)

Evaluate: x→2lim​[x2−4x3−4x2+4x​]

If f(x)=x3−x31​, show that f(x)+f(x1​)=0

The value of x→2lim​x−2(x4−16)​ is

If x→3lim​x−3xn−3n​=108, find the value of n.

Solve the equation:- limx→π/2​ x−π/2tan 2x​