JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

\lim \limits_ {\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}=\ldots \ldots

View answer and explanation

Correct option: A

\sqrt{2}

Option A (

\sqrt{2}

) is correct because the limit can be evaluated using L'Hopital's rule, resulting in cos(

\alpha

) + sin(

\alpha

), which evaluates to

\sqrt{2}

\alpha

\pi

/4.

More Questions from LIMITS

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\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}=

\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\}

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}

f^{\prime}(0)

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}