given 
is continuous but not differentiable at 
Solution: $f(x)=|x-3|$ As every modulus function is continuous for all real $x, f(x)$ is continuous at $x=3$. $f(x)=f(x)=\left\{\begin{array}{l} 3-x, x<0 \\ x-3, x \geq 0 \end{array}\right.$ In...
, defined as 
at 
Solution: $: \mathrm{f}$ is continuous at $x=0$ $\lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0+} f(x)$ $\lim _{x \rightarrow 0-}\left(\operatorname{asin} \frac{\pi}{2}(x+1)\right)=\lim _{x...
Solution: $f$ is continuous at $x=2$ $\begin{array}{l} \lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2+} f(x)=f(2) \\ \lim _{x \rightarrow 2-}(5)=\lim _{x \rightarrow 2+}[a x+b]=5 \\...
is continuous at 
Solution: Left Hand Limit: $\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 1-} x^{3}-3$ $=5$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} x^{2}+1$ $=5$...
is continuous at 
Solution: $\begin{array}{l} : \mathrm{Left Hand Limit}: \lim _{\mathrm{x} \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 1-} \mathrm{x}^{2}+1 \\ =2 \end{array}$...
Solution: $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^{2} \sin \frac{1}{x}$ As $\lim _{x \rightarrow 0} x^{2}=0$ and $\sin \left(\frac{1}{x}\right)$ is bounded function between $-1$ and...
is the following function 
Ans. 
Solution: $f$ is continuous at $x=\frac{\pi}{2}$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow-\frac{\pi}{r}} f(x)=f\left(\frac{\pi}{2}\right) \\ \Rightarrow \lim _{x \rightarrow \frac{\pi}{r}}...
is the following function 
is continuous at Ans. Solution: As, $f(x)$ is continuous at $x=3$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}=f(3) \\ \Rightarrow \lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{x-3}=f(3) \\...
is the following function continuous at 
Solution: As, $f(x)$ is continuous at $x=2$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow 2-} 2 x+1=\lim _{x \rightarrow 2^{*}} 3 x-1=f(2) \\ \Rightarrow \lim _{x \rightarrow 2^{-}} 2 x+1=f(2)...
for which 
Solution: As, $f(x)$ is continuous at $x=0$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow-1} \frac{x^{2}-2 x-3}{x+1}=f(0) \\ \Rightarrow \lim _{x \rightarrow-1} \frac{(x-3)(x+1)}{x+1}=\lambda...
for which 
Solution: As, $f(x)$ is continuous at $x=0$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 2 \mathrm{x}}{5 \mathrm{x}}=\mathrm{f}(0) \\ \Rightarrow \frac{1}{5} \lim _{\mathrm{x}...
Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=-x \\ =0 \end{array}$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2 *} x$ $=0$...
Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 2-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{-}} 2 \mathrm{x}$ $=4$ Right Hand Limit: $\lim _{\mathrm{x} \rightarrow 2^{-}}...
Solution: $\lim _{x \rightarrow 0} \sin \frac{1}{x}=0$ $\sin _{\mathrm{x}}^{1}$ is bounded function between $-1$ and $+1$ Also, $f(0)=0$ As, $\lim _{x \rightarrow 0} f(x)=f(0)$ As a result,...
is discontinuous at x=0Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 0-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0-2} \frac{1}{2}(\mathrm{x}-|\mathrm{x}|)$ $=\lim _{x \rightarrow 0-2}...
Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow \mathrm{a}-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \mathrm{a}-} \frac{|\mathrm{x}-\mathrm{a}|}{\mathrm{x}-\mathrm{a}}$ $=\lim...
Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0-} \cos x \\ =1 \end{array}$ Right Hand Limit: $\lim _{\mathrm{x} \rightarrow 0_{-}}...
is continuous at Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 2-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{x}-1$ $=1$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim...
Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1-} 5 x-4 \\ =1 \end{array}$ Right Hand Limit: $\lim _{\mathrm{x} \rightarrow 1^{*}}...
Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0-} 3-x \\ =3 \end{array}$ $\begin{array}{l} \text { Right Hand Limit: } \lim _{x...
is discontinuous at Solution : Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 2-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2-} 2+\mathrm{x}$ $=4$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim...
Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0-} \frac{1-\operatorname{cosx}}{x^{2}} \\ =\lim _{x \rightarrow 0-} \frac{2 \sin ^{2}...
Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 0-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0-} \frac{\sin 3 \mathrm{x}}{\mathrm{x}}$ $=3$ $\left[\lim _{x \rightarrow a}...
Solution: Left Hand Limit: $\lim _{x \rightarrow 5-} f(x)=\lim _{x \rightarrow 5-} \frac{x^{2}-25}{x-5}$ $=\lim _{x \rightarrow 5-} \frac{(x+5)(x-5)}{x-5}$ [By middle term splitting]...
is continuous at Solution: L.H.L.: $\lim _{\mathrm{x} \rightarrow 1-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 1^{-}} \mathrm{x}^{2}+3 \mathrm{x}+4$ $=7$ R.H.L.: $\lim _{x \rightarrow 1^{-}} f(x)=\lim...
is continues at Solution: L.H.L.: $\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2-} x^{2}$ $=4$ R.H.L.: $\lim _{\mathrm{x} \rightarrow 2^{*}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{*}}...