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# Sandwich Theorem - Practice Questions & MCQ

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

## Quick Facts

• 62 Questions around this concept.

## Solve by difficulty

$\text { If } \lim _{x \rightarrow 0} \frac{a e^x+b \cos x+c e}{e^{2 x}-2 e^x+1}=4 \text {, then }$:

:

Which of the following is true for the real valued functions $f(x),h(x),g(x)$

$\lim _{x \rightarrow a} f(x)=L, \lim _{x \rightarrow-a} g(x)=L, \lim _{x \rightarrow-a} f(x)=N, \lim _{x \rightarrow a} g(x)=N$?

The integer $\mathrm{n}$ for which $\mathrm{\lim _{x \rightarrow 0}\left\{\frac{(\cos x-1)\left(\cos x-e^2\right)}{x^n}\right\}}$ is a finite, non-zero real number, is

$\mathrm{\lim _{n \rightarrow \infty}\left[\sqrt{\left(n^2+n+1\right)}-\left[\sqrt{\left(n^2+n+1\right)}\right]\right)}$
where [ ] denotes the greatest integer function:

$\mathrm{\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r}{\left(n^2+n+r\right)} \text { equals : }}$

$\mathrm{ Assume\, \, that \lim _{\theta \rightarrow-1} f(\theta) exists and \frac{\theta^2+\theta-2}{\theta+3} \leq \frac{f(\theta)}{\theta^2} \leq \frac{\theta^2+2 \theta-1}{\theta+3} }$
$\mathrm{ holds\, \, for \, \, certain\, \, interval\, \, containing\, \, the \, \, point \, \, \theta=-1 then \lim _{\theta \rightarrow-1} f(\theta) : }$

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$\text{ Evaluate :}\mathrm{\lim _{n \rightarrow \infty} \frac{1}{1+n^2}+\frac{2}{2+n^2}+\ldots .+\frac{n}{n+n^2}}$.

$\text{ Evaluate :}\mathrm{\lim _{x \rightarrow \infty} \frac{x+7 \sin x}{-2 x+13}}$  using Sandwich Theorem.

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$\mathrm{\text { If }[x] \text { denotes the integral part of } x \text {, then } \lim _{x \rightarrow \infty} \frac{\log _e[x]}{x}=}$.

## Concepts Covered - 1

Sandwich Theorem

Sandwich Theorem

Sandwich theorem is also known as squeeze play theorem. It is typically used to find the limit of a function via comparison with two other functions whose limits are known or are easily calculated.

Sandwich Theorem

Let f(x), g(x) and h(x) be real functions such that f (x) ≤ g( x) ≤ h(x) for all x in neighbourhood of x = a. If $\lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x), \text { then } \lim _{x \rightarrow a} g(x)=l .$

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Sandwich Theorem

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## Books

### Reference Books

#### Sandwich Theorem

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.7

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