Question

Show that the function f(x)={x+λ,x<1λx2+1,x≥1 is continuous function, regardless of the choice of λ∈R

Answer 1

Answer:

I hope this helps you.

For x ∈ ( -∞, 1 ), the function is given by the linear formula f(x) = x + λ and so is continuous on that interval.

For x ∈ ( 1, ∞ ), the function is given by the polynomial f(x) = λx² + 1 and so is continuous on that interval.

All that remains is to show that the function is continuous at x=1.

Need:

• limit to exist there
• function to be defined there
• the values in the two above points should be equal

Limit exists

Left and right hand limits exist.  If they're equal, we're done.

For the left hand limit (i.e. limit as x approaches 1 from the left), since we're looking at values of x < 1, we need to use the appropriate formula.  So:

left hand limit

$\displaystyle=\lim_{x\rightarrow1^-}f(x)=\lim_{x\rightarrow1^-}x+\lambda=1+\lambda$

Similarly,

right hand limit

$\displaystyle=\lim_{x\rightarrow1^+}f(x)=\lim_{x\rightarrow1^+}\lambda x^2+1=\lambda+1$

As these two limits exist and are equal to λ+1, we can conclude that the limit exists and is equal to λ+1.

Function defined there

The definition of f(x) states that at x=1 we should use the second formula.  So...

f(1) = λ 1² + 1 = λ + 1

Limit and value are equal

Since $\lim_{x\rightarrow 1}f(x) = f(1)$, it follows that f is continuous at x=1.

In view of the comments at the beginning, it follows that f is continuous on all of R.