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Sandwich Theorem - Practice Questions & MCQ

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

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  • 62 Questions around this concept.

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\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 }:

\text { Evaluate: } \lim _{x \rightarrow 0}\left(\frac{1^x+2^x+3^x+\ldots+n^x}{n}\right)^{a / x}:

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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