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Examine the continuity of the function $f(x)=x^3+2x^2-1$ at $x=1$

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  • A function $y=f(x)$ is said to be continuous at $x=a$,if $\lim\limits_{\large x\to a}f(x)=f(a)$ and left hand limit is equal to the right hand limit.
Step 1:
Let the function be $f(x)=x^3+2x^2-1$
We are asked to check the continuity at $x=1$.
$\lim\limits_{\large x\to 0^-}f(x)=\lim\limits_{\large h\to 0}(1-h)$
$\qquad\qquad=\lim\limits_{\large h\to 0}(1-h)^3+2(1-h)^2-1$
Step 2:
On expanding and simplifying we get,
$\qquad\qquad=\lim\limits_{\large h\to 0}1-3h^2+3h-h^3+2-4h+h^2-1$
On applying limits we get,
$\qquad\qquad=2$
Step 3:
$\lim\limits_{\large x\to 0^+}f(x)=\lim\limits_{\large h\to 0}(1+h)$
$\qquad\qquad=\lim\limits_{\large h\to 0}(1+h)^3+2(1+h)^2-1$
$\qquad\qquad=\lim\limits_{\large h\to 0}1+3h^2+3h+h^3+2+4h+h^2-1$
On applying limits we get,
$\qquad\qquad=2$
Step 4:
Since the left hand limit=right hand limit.
Therefore the function is continuous at $x=1$
answered Jun 25, 2013 by sreemathi.v
 

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