Q

# 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 ?

Views

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.

Exams
Articles
Questions