### Home > CALC > Chapter Ch4 > Lesson 4.4.1 > Problem4-126

4-126.

Differentiability implies continuity. If you are checking for differentiability, you also need to check for continuity.

To check for continuity:

$\text{Verify that:}\lim_{x\rightarrow 2^{-}}g(x)=\lim_{x\rightarrow 2^{+}}g(x)$

$\text{And verify that:}\lim_{x\rightarrow 2}g(x)=g(2)$

$\lim_{x\rightarrow 2^{-}}g(x)=\lim_{x\rightarrow 2^{-}}(x-1)^{2}=1$

$\lim_{x\rightarrow 2^{+}}g(x)=\lim_{x\rightarrow 2^{+}}2\text{sin}(x-2)+1=1$

g(2) = 2(sin(2 − 2)) + 1 =1

$\lim_{x\rightarrow 2^{+}}g(x)=g(2)$

g(x) is continuous at x = 2.

To check for differentiability:

$\text{Verify that:}\lim_{x\rightarrow 2^{-}}g'(x)=\lim_{x\rightarrow 2^{+}}g'(x)=g'(2)$

$\lim_{x\rightarrow 2^{-}}g'(x)=\lim_{x\rightarrow 2^{-}}(2x-2)=2$

$\lim_{x\rightarrow 2^{+}}g'(x)=\lim_{x\rightarrow 2^{+}}2\text{cos}(x-2)=2$

g'(2) = 2cos(2 - 2) = 2 ............................

Since both g(x) and g'(x) are continuous at x = 2, g(x) is differentiable at x = 2.