Q: \$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}\$=

\$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}\$\$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { x\dfrac { 2\tan { x } }{ 1-\tan ^{ 2 }{ x } } -2x\tan x }{\left(1-\dfrac { 1-\tan ^{ 2 }{ x } }{ 1+\tan ^{ 2 }{ x } } \right)^{ 2 } } } \$\$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x } }{ 4\tan ^{ 4 }{ x } }}\$\$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x } } } \$\$=\displaystyle \dfrac { 1 }{ 2 } \$

Q: lf \$ \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) \$ is finite then the value of \$a,b\$ respectively are

\$\lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) \$As \${ x\rightarrow 0 }\$, denominator tends to 0, so the numerator also tends to 0.\$\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0\$at \$x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1\$\$\Rightarrow 1+a+b=0\$\$\Rightarrow a+b=-1\$Now, put \$a = -4 \$\$ \Rightarrow b = 3\$Option C satisfies above equation.

Q: \$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=\$

\$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})\$\$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 } } ) } \$It is of the form \$\displaystyle \dfrac{0}{0}\$, so applying L-Hospital's rule \$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) } } \$\$=\displaystyle \dfrac{2}{\pi}\$

Q: \$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}\$=

\$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=\lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin 2x}\$\$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}\$...... As\$\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}\$

Q: \$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=\$

Using, \$1 - cos 2x = 2{sin}^{2} x \$,The expression transforms to,\$lim _{ x \rightarrow 0 } \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x} \$Rewriting the expression in a different form,\$lim _{ x \rightarrow 0 } \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3} \$Therefore, the limit to the expression is \$ \dfrac{10}{3} \$Hence, option 'A' is correct.

Q: \$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}\$=

\$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { (\dfrac { \pi }{ 2 } -x)\sec x }{ \csc { x } } \$\$\displaystyle=\lim _{ h\rightarrow 0 } \dfrac { (\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 2 } +h)\sec { (\dfrac { \pi }{ 2 } -h) } }{ \csc { (\dfrac { \pi }{ 2 } -h } ) }\$\$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\csc { h } }{ \sec { h } } \$\$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\cos { h } }{ \sin { h } } =1\$

Q: \$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=\$

\$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}\$\$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }\$\$ =\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) } \$\$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}\$

Q: The value of \$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } \$ is:

\$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] } \$\$\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] } \$\$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right) \$ ....... \$[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]\$\$\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }\$ ..... \$[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]\$\$=\dfrac { \sin { 2\beta } }{ 2\beta } \$

Q: Evaluate: \$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}\$

Given,\$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}\$\$\sec x=\dfrac{1}{\cos x}\$\$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}\$\$=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}\$\$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}\$\$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}\$as \$\cos 0=1\$ and \$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1\$\$=\dfrac{3x}{2x}=\dfrac{3}{2}\$

Q: \$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=\$

We simplify the given expression as,\$\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)} \$Let \$x - sin x = y\$As \$ x \rightarrow 0 \text{ so does y } \rightarrow 0 \$Hence, the question transforms into\$(\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow 0}\dfrac{{e}^{y} - 1}{y}) \$\$= \dfrac{1}{2} \times 1 \$\$= \dfrac{1}{2} \$

Question 1

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$=

Accepted Answer

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { x\dfrac { 2\tan { x }  }{ 1-\tan ^{ 2 }{ x }  } -2x\tan  x }{\left(1-\dfrac { 1-\tan ^{ 2 }{ x }  }{ 1+\tan ^{ 2 }{ x }  } \right)^{ 2 } }  } $$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x }  }{ 4\tan ^{ 4 }{ x }  }}$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x }  } }  $$=\displaystyle \dfrac { 1 }{ 2 } $

Question 2

lf $ \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle  \frac { \cos  4x+a\cos  2x+b }{ x^{ 4 } }  \right) $ is finite then the value of $a,b$ respectively are

Accepted Answer

$\lim _{ x\rightarrow 0 } \left(\displaystyle  \frac { \cos  4x+a\cos  2x+b }{ x^{ 4 } }  \right) $As ${ x\rightarrow 0 }$, denominator tends to 0, so the numerator also tends to 0.$\Rightarrow \lim _{ x\rightarrow 0 } \cos  4x+a\cos  2x+b=0$at $x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1$$\Rightarrow 1+a+b=0$$\Rightarrow a+b=-1$Now, put $a = -4 $$ \Rightarrow b = 3$Option C satisfies above equation.

Question 3

$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=$

Accepted Answer

$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 }  } ) } $It is of the form $\displaystyle \dfrac{0}{0}$, so applying L-Hospital's rule $=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi  }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) }  } $$=\displaystyle \dfrac{2}{\pi}$

Question 4

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}$=

Accepted Answer

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=\lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin 2x}$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}$...... As$\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}$

Question 5

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=$

Accepted Answer

Using, $1 - cos 2x = 2{sin}^{2} x $,The expression transforms to,$lim _{ x \rightarrow 0  }        \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x} $Rewriting the expression in a different form,$lim _{ x \rightarrow 0 }          \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3} $Therefore, the limit to the expression is $ \dfrac{10}{3} $Hence, option 'A' is correct.

Question 6

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}$=

Accepted Answer

$\displaystyle \lim _{ x\rightarrow \dfrac { \pi  }{ 2 }  } \dfrac { (\dfrac { \pi  }{ 2 } -x)\sec  x }{ \csc { x }  } $$\displaystyle=\lim _{ h\rightarrow 0 } \dfrac { (\dfrac { \pi  }{ 2 } -\dfrac { \pi  }{ 2 } +h)\sec { (\dfrac { \pi  }{ 2 } -h) }  }{ \csc { (\dfrac { \pi  }{ 2 } -h } ) }$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\csc { h }  }{ \sec { h }  } $$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\cos { h }  }{ \sin { h }  } =1$

Question 7

$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=$

Accepted Answer

$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}$$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 }  }  }{ x\log  (1+x) }$$ =\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 }  }  }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 }  }{\displaystyle \dfrac { 1 }{ x } \log  (1+x) } $$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 }  }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}$

Question 8

The value of $\displaystyle \lim _{ \alpha \rightarrow \beta  }{ \left[ \frac { \sin ^{ 2 }{ \alpha  } -\sin ^{ 2 }{ \beta  }  }{ { \alpha  }^{ 2 }-{ \beta  }^{ 2 } }  \right]  } $ is:

Accepted Answer

$\displaystyle \lim _{ \alpha \rightarrow \beta  }{ \left[ \frac { \sin ^{ 2 }{ \alpha  } -\sin ^{ 2 }{ \beta  }  }{ { \alpha  }^{ 2 }-{ \beta  }^{ 2 } }  \right]  } =\lim _{ \alpha \rightarrow \beta  }{ \left[ \frac { \sin { \left( \alpha -\beta  \right)  } \sin { \left( \alpha +\beta  \right)  }  }{ \left( \alpha -\beta  \right) \left( \alpha +\beta  \right)  }  \right]  } $$\displaystyle =\lim _{ \alpha \rightarrow \beta  }{ \left[ \frac { \sin { \left( \alpha -\beta  \right)  }  }{ \left( \alpha -\beta  \right)  }  \right]  } \times \lim _{ \alpha \rightarrow \beta  }{ \left[ \frac { \sin { \left( \alpha +\beta  \right)  }  }{ \left( \alpha +\beta  \right)  }  \right]  } $$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta  \right)  }  }{ \left( \alpha -\beta  \right)  }  \right]  } .\left( \frac { \sin { 2\beta  }  }{ 2\beta  }  \right) $ ....... $[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]$$\displaystyle =1.\frac { \sin { 2\beta  }  }{ 2\beta  }$ ..... $[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]$$=\dfrac { \sin { 2\beta  }  }{ 2\beta  } $

Question 9

Evaluate: $\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$

Accepted Answer

Given,$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$\sec x=\dfrac{1}{\cos x}$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}$$=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$as $\cos 0=1$ and $\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$=\dfrac{3x}{2x}=\dfrac{3}{2}$

Question 10

$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=$

Accepted Answer

We simplify the given expression as,$\displaystyle\lim_{x \rightarrow 0}     \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)} $Let $x - sin x = y$As $ x \rightarrow 0 \text{ so  does  y } \rightarrow 0 $Hence, the question transforms into$(\displaystyle\lim_{x \rightarrow 0}  \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow  0}\dfrac{{e}^{y} - 1}{y}) $$=  \dfrac{1}{2}  \times 1 $$=  \dfrac{1}{2} $

Limits And Derivatives

Maths
5256 Views

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Limits And Derivatives

Maths 5256 Views

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