JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Application +4 | -1

Let

f(x)=\left\{\begin{array}{cc}\frac{\sin 4 x}{\log (1+3 x)}, & x \neq 0 \\ A+1, & x=0\end{array}\right.

The value of

A

for

f

to be continuous at

x=0

is

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

) is correct because for

f

to be continuous at

x=0

,

A+1 = \lim_{x \rightarrow 0} \frac{\sin 4x}{\log(1+3x)} = \frac{4}{3}

, which gives

A = \frac{1}{3}

.

More Questions from LIMITS

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}

, then the value of

f^{\prime}(0)

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