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=$\pi$