Aspire Faculty ID #18303 · Topic: AMU MCA 2017 · Just now
AMU MCA 2017

Let $f(x)$ be a function defined by $f(x)=\begin{cases}4x-5,\ x\le2 \ x-\lambda,\ x>2 \end{cases}$ If $\lim_{x\to2} f(x)$ exists, then the value of $\lambda$ is

Solution

For limit to exist,

$ \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x) $

$ \lim_{x\to2^-} f(x)=4(2)-5=3 $

$ \lim_{x\to2^+} f(x)=2-\lambda $

Equating,

$ 3=2-\lambda $

$ \Rightarrow \lambda=-1 $

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