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Limit - Practice Questions & MCQ

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

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  • 31 Questions around this concept.

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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).

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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:

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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

Limit

Limit

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.

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Limit

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