Basic maths-calculus-Q2

+1 vote

lim_{x\rightarrow-2} \frac{(x+3)|x+2|}{x+2}

The value is

(A) 1

(B) -1

(C) 0

(D) does not exist

asked May 19 in Basic Maths by gbeditor (44,560 points)
reshown May 24 by gbeditor

3 Answers

0 votes
Best answer

Right limit : 

x \rightarrow -2^+

x = -2+h; h \rightarrow 0

lim_{x\rightarrow-2} \frac{(x+3)|x+2|}{x+2}

\lim_{h \rightarrow 0} \frac{(h+1)(|h|)}{h}

Since we are finding right limit, hence, x \rightarrow -2^+   ;  x = -2+h; h \rightarrow 0; Since,h is positive,so, |h| = h

\lim_{h \rightarrow 0} \frac{(h+1)(|h|)}{h} = 1

Left limit : 

x \rightarrow -2^-

x = -2-h; h \rightarrow 0

lim_{x\rightarrow-2} \frac{(x+3)|x+2|}{x+2}

\lim_{h \rightarrow 0} \frac{(-h+1)(|-h|)}{-h}

Since, h is positive, so, |-h| = h

\lim_{h \rightarrow 0} \frac{(-h+1)(|-h|)}{-h} = -1

So, left limit is Not same as right limit, hence, limit does not exist.

answered May 27 by deepak-gatebook (112,760 points)
0 votes
Can we have the solutions? Linear algebra solutions are also missing for some question...
answered May 26 by tsabhineet21 (7,690 points)
Most of the answers have been updated in detail. Check out now.Other will be updated shortly.
0 votes
|x+2| = x+2 for x+2>=0 i.e. x>=-2

|x+2|=-(x+2) for x+2<0 i.e. x<-2

Clearly, left hand limit and right hand limit of f(x) at x=-2 are not equal. Hence (D)
answered May 27 by rahulfabledpro21 (180 points)