Question

Is the function defined by \(f(x)=x^2-sin\,x+5\) continuous at \(x=p\) ?

Answer 1

The given function is f(x)=x²-sinx+5
It is evident that f is defined at x=p 
At x=\(\pi\),f(x)=f(\(\pi\))=\(\pi\)²-sin\(\pi\)+5=\(\pi\)²-0+5=\(\pi\)²+5
Consider \(\lim_{x\rightarrow \pi}\)f(x)=\(\lim_{x\rightarrow \pi}\)(x²-sinx+5)
put x=\(\pi\)+h
If x\(\rightarrow\)\(\pi\), then it is evident that h\(\rightarrow\)0
∴\(\lim_{x\rightarrow \pi}\)f(x)=\(\lim_{x\rightarrow \pi}\)(x²-sinx+5)
=\(\lim_{h\rightarrow 0}\)[(\(\pi\)+h)²-sin(\(\pi\)+h)+5]
=\(\lim_{h\rightarrow 0}\)(\(\pi\)+h)²-\(\lim_{h\rightarrow 0}\)sin(\(\pi\)+h)+\(\lim_{h\rightarrow 0}\)5
=(\(\pi\)+0)²-\(\lim_{h\rightarrow 0}\)[sin\(\pi\)cosh+cos\(\pi\)sinh]|+5
=\(\pi\)²-\(\lim_{h\rightarrow 0}\) sin\(\pi\)cosh-cos\(\pi\)sinh+5
=\(\pi\)²-sin\(\pi\)cos0-cos\(\pi\)sin0+5
=\(\pi\)²-0x1-(-1)x0+5
=\(\pi\)²+5
∴\(\lim_{x\rightarrow \pi}\) f(x)=f(\(\pi\))
Therefore, the given function f is continuous at x=\(\pi\)