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

Answers (1)
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Given,

f (x) = \left\{\begin{matrix} mx^2 + n & x < 0 \\ nx + m & 0 \leq x \leq 1 \\ nx^3+m & x > 1 \end{matrix}\righ t.

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.

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