Limit – JEE Advanced Previous Year Questions with Solutions

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Q. Let $\mathrm{L}=\lim _{x \rightarrow 0} \frac{\mathrm{a}-\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}-\frac{\mathrm{x}^{2}}{4}}{\mathrm{x}^{4}}, \mathrm{a}>0 .$ If $\mathrm{L}$ is finite, then $:-$

(A) a = 2

(B) a = 1

(C) $\mathrm{L}=\frac{1}{64}$

(D) $\mathrm{L}=\frac{1}{32}$

[JEE 2009, 4]

Sol. (A,C)

$\mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}\right)^{\frac{1}{2}}-\frac{\mathrm{x}^{2}}{4} \quad \mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{2 \mathrm{a}^{2}}-\frac{1}{8} \frac{\mathrm{x}^{4}}{\mathrm{a}^{4}}\right)-\frac{\mathrm{x}^{2}}{4}$

$\mathrm{a}=2,\left(\mathrm{coefficient} \text { of } \mathrm{x}^{2}=0\right)$

$\therefore \mathrm{L}=\frac{1}{64}$

Q. If $\lim _{x \rightarrow 0}\left[1+x \ell n\left(1+b^{2}\right)\right]^{\frac{1}{x}}=2 b \sin ^{2} \theta, b>0$ and $\theta \in(-\pi, \pi],$ then the value of $\theta$ is-

[JEE 2011, 3M, –1M]

Sol. (D)

Q. If $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4,$ then $-$

(A) a = 1, b = 4 (B) a = 1, b = –4] (C) a = 2, b = –3 (D) a = 2, b = 3

[JEE 2012, 3M, –1M]

Sol. (B)

Q. Let $\alpha(\mathrm{a})$ and $\beta(\mathrm{a})$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a>-1 .$ Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

(A) $-\frac{5}{2}$ and 1

(B) $-\frac{1}{2}$ and $-1$

(C) $-\frac{7}{2}$ and 2

(D) $-\frac{9}{2}$ and 3

[JEE 2012, 3M, –1M]

Sol. (B)

Q. The largest value of the non-negative integer a for which $\lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}$ is

[JEE(Advanced)-2014, 3]

Sol. 0

Q. Let $\alpha, \beta \in \mathrm{R}$ be such that $\lim _{x \rightarrow 0} \frac{x^{2} \sin (\beta x)}{\alpha x-\sin x}=1 .$ Then $6(\alpha+\beta)$ equals


Sol. 7

Q. Let $\mathrm{f}(\mathrm{x})=\frac{1-\mathrm{x}(1+|1-\mathrm{x}|)}{|1-\mathrm{x}|} \cos \left(\frac{1}{1-\mathrm{x}}\right)$ for $\mathrm{x} \neq 1 .$ Then


Sol. (A,C)

Q. For any positive integer n, define $f_{\mathrm{n}}:(0, \infty) \rightarrow \square$ as $f_{\mathrm{n}}(\mathrm{x})=\sum_{\mathrm{j}=1}^{\mathrm{n}} \tan ^{-1}\left(\frac{1}{1+(\mathrm{x}+\mathrm{j})(\mathrm{x}+\mathrm{j}-1)}\right)$ for all $\mathrm{x} \in(0, \infty)$

(Here, the inverse trigonometric function $\tan ^{-1} \mathrm{x}$ assume values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$ )

Then, which of the following statement(s) is (are) TRUE?

(A) $\sum_{j=1}^{5} \tan ^{2}\left(f_{j}(0)\right)=55$

(B) $\sum_{j=1}^{10}\left(1+f_{j}^{\prime}(0)\right) \sec ^{2}\left(f_{j}(0)\right)=10$

(C) For any fixed positive integer $n, \lim _{x \rightarrow \infty} \tan \left(f_{\mathrm{n}}(x)\right)=\frac{1}{n}$

(D) For any fixed positive integer $n, \operatorname{limsec}_{x \rightarrow \infty} \operatorname{ec}^{2}\left(f_{\mathrm{n}}(x)\right)=1$


Sol. (D)

Q. For each positive integer $n,$ let $y_{n}=\frac{1}{n}(n+1)(n+2) \ldots(n+n)^{1 / n}$ For $x \in \square,$ let $[x]$ be the greatest integer less than or equal to $x$. If $\lim _{n \rightarrow \infty} y_{n}=L,$ then the value of $[L]$ is


Sol. 1


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  • August 10, 2020 at 1:51 pm

    questions are not clear