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31. Given, $\underset{x\to 1}{lim}\frac{f(x)-2}{x^2-1}=\pi$

$\frac{\underset{x\to 1}{lim}f(x)-2}{\underset{x\to 1}{lim}{x}^2-1}=\pi$

$\Rightarrow \underset{x\to 1^{}}{lim}[f(x)-2]=\pi \underset{x\to 1}{lim}\left(x^2-1\right)$

$\Rightarrow \underset{x\to 1}{lim}f(x)-\underset{x\to 1}{lim}2=\pi \left(1^2-1\right)$

$\Rightarrow \underset{x\to 1}{lim}f(x)-2=0.$

$\Rightarrow \underset{x\to 1}{lim}f(x)=2$

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Jaipur Noida University provides various scholarships in the form of fee concession, award money, financial assistance etc. The type of scholarships JNU Jaipur provides to BBA students and other course students are mentioned below:

30. Given, $f(x)=\{\begin{array}{ccc}\hfill \left|x\right|+1,\hfill & \hfill x<0\hfill & \hfill 0,\hfill \\ \hfill x=0\hfill & \hfill \left|x\right|-1,\hfill & \hfill x>0\hfill \end{array}\underset{x\to a}{lim}f(x)=?$

As | x | = $\{\begin{array}{l}-x,x<0\\ x,x>0\end{array}$

We can rewrite f (x) = $\begin{array}{l}\{\begin{array}{l}-x+1,x<0\\ 0,x=0\end{array}\\ x-1,x>0.\end{array}$

Case 1: when a<0,

$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}(-x+1)=-a+1$

So, $\underset{x\to a}{lim}f(x)$ = exist such that a< 0

Case II when a> 0,

$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}(x-1)=a-1$

So, $\underset{x\to a}{lim}f(x)$ exist such that a>0.

Case III when a = 0.

L.H.L = $\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{-}}{lim}(-x+1)=\underset{x\to 0^{-}}{lim}(-x+1)=1$

R.H.L = $\underset{x\to a^{+}}{lim}f(x)=\underset{x\to a^{+}}{lim}(x-1)=\underset{x\to 0^{+}}{lim}(x-1)=-1$

Thus, $\underset{x\to a^{-}}{lim}f(x)\ne \underset{x\to a^{+}}{lim}f(x)$

So, $\underset{x\to a}{lim}f(x)$ does not exist at a = 0.

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No, SRK Institute of Technology M.Tech admissions are generally not merit-based. Students need to complete graduation from a recognised university in a relevant field. Further, at the time admission, students must provide valid score obtained in GATE/ PGECET. Hence, no M.Tech admissions at SRKIT Vijayawada are not merit-based. For more information, students can visit the official website of the institute. 

Cochin University of Science and Technology is one of the top Engineering institutions in the country. The university builds on the state reputation of friendliness and focus on all-round development. Students can choose from a diversity of courses like BTech, MTech, MBA, MSc, etc., with different specializations. CUSAT is located in the quaint and serene setting of Kochi with a number of state-of-the-art amenities available at fingertips.

Students looking for quality technical education can select any of the various BTech courses at CUSAT. In order to be eligible for Cochin University of Science and Technology BTech, students must ensure qualification from their Higher Secondary and previous qualifying examinations.

Additionally, the university also requires the relevant entrance-test for a large number of specialised courses. These entrance tests include CUSAT CAT,  CUSAT DAT,  CAT,  CMAT,  KMAT,  KMAT Kerala, etc.

28. Given, f (x) = $\begin{array}{l}\{\begin{array}{l}a+bx,x<1\\ 4,x=1\end{array}\\ b-ax,x>1\end{array}$

Since we need $\underset{x\to 1}{lim}f(x)$ we need,

LHL =

$\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}(a+bx)$ = a + b × 1 = a + b

and RHL =

$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}(b-ax)$ = b - a × 1 = b - a

Given,  $\underset{x\to 1}{lim}f(x)=f(1):$ we have the following equations

a + b = 4 ____ (1)

b - a = 4 ____ (2)

Adding (1) and (2) we get,

2b = 8

b = 4

Putting b = 4 in (1) we get

a + 4 = 4

a = 0

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27. $\underset{x\to 5}{lim}f(x)=\underset{x\to 5}{lim}\left|x\right|-5$

= | 5 |$-\mathrm{}$ 5

= 5 $-\mathrm{}$5

= 0

CUSAT has consistently remained one of the top-ranked public universities in the state of Kerala, with a ranking of #51 in the overall category. The university is also put in the 10th place for state universities category. Students can choose from a host of courses like BTech, MTech, MBA, etc. in different specializations.

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26. Given, f (x) $\{\begin{array}{ll}\frac{x}{\left|x\right|},\hfill & x\ne 0\hfill \\ 0,\hfill & x=0\hfill \end{array}$

L.H.S = $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{x}{-x}=\underset{x\to 0^{-}}{lim}-1=-1$

R.H.L $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}\frac{x}{x}=\underset{x\to 0^{+}}{lim}1=1$

Thus,  $\underset{x\to 0^{-}}{lim}f(x)\ne \underset{x\to 0^{+}}{lim}f(x)$

i e,  $\underset{x\to 0}{lim}f(x)$ does not exist.

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25. Given f (x) = $\{\begin{array}{cc}\hfill \frac{\left|x\right|}{x},x\ne 0\hfill & \hfill 0,x=0\hfill \end{array},\underset{x\to 0}{lim}f(x)=?$

We know that, $\left|x\right|=\{\begin{array}{cc}\hfill x,\hfill & \hfill x\ge 0\hfill \\ \hfill -x,\hfill & \hfill x<0\hfill \end{array}$

Now,

L.H.L = $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{-x}{x}=\underset{x\to 0^{-}}{lim}-1=-1$

and R.H.L = $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}\frac{x}{x}=\underset{x\to 0^{+}}{lim}1=1$

Thus, $\underset{x\to 0^{-}}{lim}f(x)\ne \underset{x\to 0^{+}}{lim}f(x)$

i e, $\underset{x\to 0}{lim}f(x)$ does not exist.

24. Given $f(x)=\{\begin{array}{cc}\hfill x^2-1,x\le 1\hfill & \hfill -x^2-1,x>1\hfill \end{array},\underset{x\to 1}{lim}f(x)=?$

Now, L.H.L = $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}\left(x^2-1\right)$

1²- 1 = 0

And R.H.L = $\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}\left(-x^2-1\right)$  (1)2 1 = 1 = 2.

Thus,  $\underset{x\to 1^{-}}{lim}f(x)\ne \underset{x\to 1^{+}}{lim}f(x)$

So,  $\underset{x\to 1}{lim}f(x)$ does not exist.

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23. Given f (x) $\{\begin{array}{ll}2x+3,\hfill & x\le 0\hfill \\ 3(x+1),\hfill & x>0\hfill \end{array}\}$

for $\underset{x\to 0}{lim}f(x),$

left hand limit, L.H.S = $\underset{x\to 0^{}}{lim}f(x)$ = $\underset{x\to 0^{}}{lim}(2x+3)$

= 2 0 + 3 = 3.

Right hand limit, R.H.L = $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}3(x+1)$

= (0 + 1) = 3 1 = 3.

Thus, $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{-}}{lim}f(x)=3$

For $\underset{x\to 1}{lim}f(x),$

L.H.L = $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}3(x+1)=3(1+1)=3\times 2=6$

R.H.L = $\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}3(x+1)=3(1+1)=3\times 2=6.$

Thus, $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1}{lim}f(x)=6.$