MHT CET · Maths · Limits
The value of \(\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}\) is .
- A 1
- B -1
- C 0
- D does not exist.
Answer & Solution
Correct Answer
(D) does not exist.
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { L.H.L. }=\lim _{x \rightarrow 0^{-}} \frac{x}{|x|+x^2} \\
& =\lim _{x \rightarrow 0^{-}} \frac{x}{-x+x^2} \\
& =\lim _{x \rightarrow 0^{-}} \frac{1}{x-1}=-1 \\
& \text { R.H.L. }=\lim _{x \rightarrow 0^{+}} \frac{x}{|x|+x^2} \\
& =\lim _{x \rightarrow 0^{+}} \frac{x}{x+x^2} \\
& =\lim _{x \rightarrow 0^{+}} \frac{1}{1+x}=1 \\
& =\text { L.H.L. } \neq \text { R.H.L. }
\end{aligned}\)
\(\therefore \quad\) Limit does not exist.
& \text { L.H.L. }=\lim _{x \rightarrow 0^{-}} \frac{x}{|x|+x^2} \\
& =\lim _{x \rightarrow 0^{-}} \frac{x}{-x+x^2} \\
& =\lim _{x \rightarrow 0^{-}} \frac{1}{x-1}=-1 \\
& \text { R.H.L. }=\lim _{x \rightarrow 0^{+}} \frac{x}{|x|+x^2} \\
& =\lim _{x \rightarrow 0^{+}} \frac{x}{x+x^2} \\
& =\lim _{x \rightarrow 0^{+}} \frac{1}{1+x}=1 \\
& =\text { L.H.L. } \neq \text { R.H.L. }
\end{aligned}\)
\(\therefore \quad\) Limit does not exist.
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- \(A\) rod \(A B, 13\) feet long moves with its ends \(A\) and \(B\) on two perpendicular lines \(O X\) and \(O Y\) respectively. When \(\mathrm{A}\) is 5 feet from \(\mathrm{O}\), it is moving away at the rate of \(3 \mathrm{feet} / \mathrm{sec}\). At this instant, B is moving at the rateMHT CET 2023 Medium
- If \(y=y(x)\) and \(\frac{2+\sin x}{y+1}\left(\frac{d y}{d x}\right)=-\cos x, y(0)=1\), then \(y\left(\frac{\pi}{2}\right)\) is equal toMHT CET 2022 Medium
- If \(y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{\cos x}}}\), then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) is given byMHT CET 2024 Medium
- The vectors \(\overline{\mathrm{p}}=\hat{\mathrm{i}}+a \hat{\mathrm{j}}+a^2 \hat{\mathrm{k}}, \overline{\mathrm{q}}=\hat{\mathrm{i}}+b \hat{\mathrm{j}}+b^2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{r}}=\hat{\mathrm{i}}+c \hat{\mathrm{j}}+c^2 \hat{\mathrm{k}}\) are non-coplanar and
\(\left|\begin{array}{lll}
a & a^2 & 1+a^3 \\
b & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|=0\)
then the value of \((a b c)\) isMHT CET 2025 Hard - MHT CET 2017 Easy
- \(\int_1^3\left[\tan ^{-1}\left(\frac{x}{x^2-1}\right)+\tan ^{-1}\left(\frac{x^2-1}{x}\right)\right] d x=\)MHT CET 2021 Medium
More PYQs from MHT CET
- \(\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x=\)MHT CET 2023 Medium
- A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by \(5 \Omega\). When a resistance of \(15 \Omega\) is used for shunting, null point moves to 300 cm . The internal resistance of the cell isMHT CET 2025 Easy
- Identify a molecule with incomplete octet from following.MHT CET 2023 Easy
- The magnitude of gravitational field at distance ' \(r_1\) ' and ' \(r_2\) ' from the centre of a uniform sphere of radius ' \(R\) ' and mass ' \(M\) ' are ' \(\mathrm{F}_1\) ' and ' \(\mathrm{F}_2\) ' respectively. The ratio ' \(\left(F_1 / F_2\right)\) ' will be (if \(r_1 \gt R\) and \(r_2 \lt R\))MHT CET 2024 Medium
- A uniform sphere has radius ' \(R\) ' and mass ' \(M\) '. The magnitude of gravitational field at distances ' \(\mathrm{r}_1\) ' and ' \(\mathrm{r}_2\) ' from the centre of the sphere are ' \(E_1\) ' and ' \(E_2\) ' respectively. The ratio \(E_1: E_2\) is ( \(r_1>R\) and \(r_2 < R\) )MHT CET 2025 Medium
- A coil of resistance \(5 \Omega\) and inductance \(4 \mathrm{H}\) is connected to a \(10 \mathrm{~V}\) battery. The energy stored in the coilMHT CET 2012 Medium