JEE MATHEMATICS MCQ: LIMITS

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f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \left( {\frac{1}{x}} \right),\;x \ne 0}\\ {0,\;x = 0} \end{array}} \right.

Then,

\lim\limits_{x \to 0} f\left( x \right)

View answer and explanation

Correct option: C

Is equal to 0

Option A (

{a^2} + {b^2} + {c^2} \ge bc + ca + ab

) is correct because it is a direct result of the AM-GM inequality or can be derived from the fact that the sum of squares is always non-negative.

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\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}

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