If f(x) = begin{cases} |x|+1, & x

Question

(A) The given function is $f(x) = \begin{cases} |x|+1, & x  0 \end{cases}$.Case $1$: $a = 0$.$\lim_{x 	o 0^-} f(x) =

Vedclass · Accepted Answer

$a 
eq 0$

Vedclass · Answer

(A) The given function is $f(x) = \begin{cases} |x|+1, & x  0 \end{cases}$.Case $1$: $a = 0$.$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (-x+1) = 1$.$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (x-1) = -1$.Since $1 
eq -1$,the limit does not exist at $a = 0$.Case $2$: $a $\lim_{x 	o a^-} f(x) = \lim_{x 	o a^-} (-x+1) = -a+1$.$\lim_{x 	o a^+} f(x) = \lim_{x 	o a^+} (-x+1) = -a+1$.Since the left-hand limit equals the right-hand limit,the limit exists for all $a Case $3$: $a > 0$.$\lim_{x 	o a^-} f(x) = \lim_{x 	o a^-} (x-1) = a-1$.$\lim_{x 	o a^+} f(x) = \lim_{x 	o a^+} (x-1) = a-1$.Since the left-hand limit equals the right-hand limit,the limit exists for all $a > 0$.Conclusion: The limit exists for all $a 
eq 0$.

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

Similar Questions

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+a x+b}-x\right]$ where $a < 0 < b$.

$\lim _{x \rightarrow 0} \frac{(1-e^x) \sin x}{x^2+x^3}$ is equal to

$\mathop {\lim}\limits_{x \to 1} \left[ {\left[ {\frac{4}{{{x^2} - {x^{ - 1}}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}}} \right]^{ - 1} + \frac{{3 \cdot ({x^4} - 1)}}{{{x^3} - {x^{ - 1}}}}} \right] = $

$\mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin ^2 x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module