$\mathop {\lim }\limits_{x \to - \infty } \,\frac{{2x - 1}}{{\sqrt {{x^2} + 2x + 1} }}$ is equal to

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to - \infty } \,\frac{{2x - 1}}{{\sqrt {{x^2} + 2x + 1} }}

is equal to

View answer and explanation

Correct option: B

- 2

Option B (-2) is correct because the limit as x approaches negative infinity simplifies to -2 after considering the sign of x in the square root.

Practice All Chapter MCQs →

\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}