Careers360 Logo
Rules of Changing Branch in IIIT - How to Change BTech Branch in IIITs

Limit - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 31 Questions around this concept.

Solve by difficulty

Find, \lim _{x \rightarrow 1} \frac{2 x^2-x-1}{x^2-1}

Let\; \mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}} n be n positive consecutive terms of an arithmetic progression. If this is its common difference, then: \lim _{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots .+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)is

\lim _{x \rightarrow 0}\left(\left(\frac{\left(1-\cos ^2(3 x)\right.}{\cos ^3(4 x)}\right)\left(\frac{\sin ^3(4 x)}{\left(\log _e(2 x+1)^5\right.}\right)\right) is equal to ___________:

Let,f(x)=\left[x^2-x\right]+|-x+[x]| where x\epsilon \mathbb{R}  and [t] denotes the greatest integer less than or equal to t. Then,f is:

\mathrm{\lim _{x \rightarrow 2+}\left(\frac{\mid x\mid^{3}}{3}-\left[\frac{x}{3}\right]\right)}, where \mathrm{[x]} denotes the greatest integer not exceeding \mathrm{x}, is equal to

\text{If}\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{(x+1)}-a x-b\right)=0,\text{find\; a and b}.

It is given that f(x)=\frac{a x+b}{x+1}, \lim _{x \rightarrow 0} f(x)=2$ and $\lim _{x \rightarrow \infty} f(x)=1$. Find $f(1).

Amity University, Noida B.Tech Admissions 2024

Asia's Only University with the Highest US & UK Accreditation

UPES B.Tech Admissions 2024

Ranked #52 among universities in India by NIRF | Highest CTC 50 LPA | 100% Placements | Last Date to Apply: 20th July

Evaluate \mathrm{\quad \lim _{x \rightarrow 0} \frac{x-\tan x}{x^{2} \tan x}}.

If  \mathrm{f(x)=\frac{\sin [x]}{[x]},[x] \neq 0,=0,[x]=0}. Where \mathrm{[x]} denotes the greatest integer less than or equal to \mathrm{x}, then \mathrm{ \lim _{x \rightarrow 0} f(x)} is equal to:

JEE Main Exam's High Scoring Chapters and Topics
This free eBook covers JEE Main important chapters & topics to study just 40% of the syllabus and score up to 100% marks in the examination.
Download EBook

If \mathrm{\lim _{x \rightarrow 0} \log _{\sec x}(x-\||a|]+\cos x)} is -\infty then a lies in the interval (where [.] denotes the greatest integer function )

Concepts Covered - 1



Consider the function f(x) = x2

Observe that as x takes values very close to 0, the value of f(x) is also close to 0. (See graph below)

We can also interpret it in another way. If we input the values of x which tend to/approach 0 (meaning close to 0, either just smaller than 0 or just larger than 0), the value of f(x) will tend to 0/approach (meaning close to 0, either just smaller than 0 or just larger than 0). Note we do NOT want to see what happens at x = 0, we just want values of x which are very close to 0.

Then we can say that, \mathrm{\lim_{x\rightarrow 0}f(x)=0} . (to be read as limit of f(x) as x tends to zero equals zero).

Similarly, when x approaches 2, the value of f(x) approaches 4, i.e.  \mathrm{\lim_{x\rightarrow 2}f(x)=4\;\;\;or\;\;\;\lim_{x\rightarrow 2}x^2=4}


\\\mathrm{In\;General,\;\;as\;\mathit{x}\rightarrow \mathit{a}\;(read\;as\;\mathit{x}\;tends\;to\;\mathit{a}),\;\mathit{f(x)\rightarrow l},\;then\;\text{then l is called limit of the function f (x) }}\\\text{which is symbolically written as}\;\lim_{x\rightarrow a}f(x)=l

Now consider the function

\\\mathrm{\quad\quad\quad\quad f(x)=\frac{x^2-6x-7}{x-7}}\\\text{We can factor the function as shown}\\\mathrm{\quad\quad\quad\quad f(x)=\frac{(x-7)(x+1)}{x-7}\;\;\;\;\text{[Cancel like factors in numerator and denominator.]}}\\\mathrm{\quad\quad\quad\quad f(x)=x+1,x\neq7}

Notice that x cannot be 7, or we would be dividing by 0, so 7 is not in the domain of the original function. In order to avoid changing the function when we simplify, we set the same condition, x ≠ 7, for the simplified function. We can represent the function graphically


What happens at x = 7 is completely different from what happens at points close to x = 7 on either side. Just observe that as the input x approaches 7 from either the left or the right, the output approaches 8. The output can get as close to 8 as we like if the input is sufficiently near 7. So we say that limit of this function at x = 7 equals 8 and it is denoted by \lim _{x \rightarrow 7} f(x)=8.

So even if the function does not exist at x = a, still the limit can exist at that point as limit is concerned only about the points close to x = a and NOT at x = a itself.

Study it with Videos


"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top