JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to \infty } \frac{{\left( {2x - 3} \right)\left( {3x - 4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}}

is equal to

View answer and explanation

Correct option: D

\frac{3}{{10}}

Option D (

\frac{3}{{10}}

) is correct because as x approaches infinity, the limit of the rational function is the ratio of the leading coefficients after applying L'Hopital's rule or dividing by the highest power of x.

More Questions from LIMITS

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

is equal to

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

The value of

\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)

is equal to If

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

, then the value of

f^{\prime}(0)

is equal to

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

equals