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JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1
If f(x)={x2−3,    2<x<32x+5,      3<x<4f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^2} - 3,\;\;2 < x < 3}\\ {2x + 5,\;\;\;3 < x < 4} \end{array}} \right., the equation whose roots are lim⁡x→3−f(x)\lim\limits_{x \to {3^ - }} f\left( x \right) and lim⁡x→3+f(x)\lim\limits_{x \to {3^ + }} f\left( x \right), is
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Correct option: C
x2−17x+66=0{x^2} - 17x + 66 = 0
Option C (x2−17x+66=0{x^2} - 17x + 66 = 0) is correct because the roots of the equation are 6 and 11, which are the left-hand and right-hand limits, respectively, and the equation is formed correctly.

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lim⁡x→0(1x2−cosec⁡2x)\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right) is equal to lim⁡x→π2cos⁡x(1−sin⁡x)2/3=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}= The value of lim⁡h→02{3sin⁡(π6+h)−cos⁡(π6+h)3h(3cos⁡ h−sin⁡ h)}\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\} is lim⁡x→∞x(e1/x−1)\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right) is equal to If f(x)=(x+1)71+x2(x2−x+1)6f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}, then the value of f′(0)f^{\prime}(0) is equal to lim⁡n→∞(1+sin⁡an)n\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n} equals
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