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  • If f (x) equals mx square plus n for x lesser than 0 . For what integers m and n does both lim x tends 0 f x and lim x tends to 1 f x exist?

32.  If

\\f(x)=\begin{bmatrix} mx^2+n\ \ ;x<0 & \\nx+m\ \ ;x \leq0\leq 1t &\\ nx^3+m\ \ ;x>1 \end{bmatrix}
. For what integers m and n does both \lim_{x \rightarrow 0}f (x) \: \: and\: \: \lim_{x \rightarrow 1} f (x)  exist ? 

Answers (1)

best_answer

Given,

\\f(x)=\begin{bmatrix} mx^2+n\ \ ;x<0 & \\nx+m\ \ ;x \leq0\leq 1t &\\ nx^3+m\ \ ;x>1 \end{bmatrix}

Case 1: Limit at x = 0 

The right-hand Limit or  Limit at x=0^+

\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} nx+m=n(0)+m=m

The left-hand limit or Limit at x=0^-

\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} mx^2+n=m(0)^2+n=n

Hence Limit will exist at x = 0 when m = n .

Case 2: Limit at x = 1

The right-hand Limit or  Limit at x=1^+

\lim_{x \rightarrow 1^+} f (x) = \lim_{x \rightarrow 1^+} nx^3+m=n(1)^3+m=n+m

The left-hand limit or Limit at x=1^-

\lim_{x \rightarrow 1^-} f (x) = \lim_{x \rightarrow 1^-} nx+m=n(1)+m=n+m

Hence Limit at 1 exists at all integers.

Posted by

Pankaj Sanodiya

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