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$f(x)=\left\{\begin{array}{1 1}mx^2+m,&x<0\\nx+m,&0 \leq x \leq 1 \\nx^3+m,&x>1 \end{array}\right.$ \[\] For what integers $m$ and $n$ does $ \lim\limits_{x \to 0} f(x)$ and $ \lim\limits_{x \to 1} f(x)$ exist?

1 Answer

The given function is
$f(x)=\left\{\begin{array}{1 1}mx^2+m,&x<0\\nx+m,&0 \leq x \leq 1 \\nx^3+m,&x>1 \end{array}\right.$
$ \lim\limits_{x \to 0^-}f(x) = \lim\limits_{x \to 0}(mx^2+n)$
$ = m(0)^2+n$
$ = n$
$ \lim\limits_{x \to 0^+}f(x) = \lim\limits_{x \to 0}(nx+m)$
$n(0)+m$
$ = m$
Thus, $ \lim\limits_{x \to 0}f(x)$ exists if $ m = n$
$ \lim\limits_{x \to 1^-}f(x) = \lim\limits_{x \to 1}(nx+m)$
$ n(1)+m$
$ = m+n$
$ \lim\limits_{x \to 1^+}f(x) = \lim\limits_{x \to 1}(nx^3+m)$
$ n(1)^3+m$
$ = m+n$
$ \therefore \lim\limits_{x \to 1^-}f(x) = \lim\limits_{x \to 1^+}f(x) = \lim\limits_{x \to 1}f(x)$
Thus, $\lim\limits_{x \to 1}f(x)$ exists for any integral value of $m$ and $n$
answered Apr 7, 2014 by thanvigandhi_1
 

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