Notes: Can someone please help me answering those problems on the attached files at the bottom?

Thursday, 15 May 2014

Can someone please help me answering those problems on the attached files at the bottom?

You are not allowed to ask multiple questions. This is a response for 3 of the several that you have posted.

# If $\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{x}+{2}}}$ , determine whether f is continuous from the left, from the right or neither for x = -2

$\displaystyle\lim_{{{x}\to-{{2}}^{+}}}{f{{\left({x}\right)}}}$

= $\displaystyle\lim_{{{x}\to-{{2}}^{+}}}\sqrt{{{x}+{2}}}$

= $\displaystyle\sqrt{{-{2}+{2}}}$

= 0

$\displaystyle\lim_{{{x}\to-{{2}}^{{-}}}}{f{{\left({x}\right)}}}$

= $\displaystyle\lim_{{{x}\to-{{2}}^{{-}}}}\sqrt{{{x}+{2}}}$

But for x < -2, $\displaystyle\sqrt{{{x}+{2}}}$ is not real as x + 2 is negative and the...

You are not allowed to ask multiple questions. This is a response for 3 of the several that you have posted.

# If $\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{x}+{2}}}$ , determine whether f is continuous from the left, from the right or neither for x = -2

$\displaystyle\lim_{{{x}\to-{{2}}^{+}}}{f{{\left({x}\right)}}}$

= $\displaystyle\lim_{{{x}\to-{{2}}^{+}}}\sqrt{{{x}+{2}}}$

= $\displaystyle\sqrt{{-{2}+{2}}}$

= 0

$\displaystyle\lim_{{{x}\to-{{2}}^{{-}}}}{f{{\left({x}\right)}}}$

= $\displaystyle\lim_{{{x}\to-{{2}}^{{-}}}}\sqrt{{{x}+{2}}}$

But for x < -2, $\displaystyle\sqrt{{{x}+{2}}}$ is not real as x + 2 is negative and the square root of a negative number is not real.

Therefore $\displaystyle\lim_{{{x}\to-{{2}}^{{-}}}}{f{{\left({x}\right)}}}\ne{f{{\left(-{2}\right)}}}$ .

The function is continuous from the right at x = -2.

# If f and g are continuous with f(3) = 5 and $\displaystyle\lim_{{{x}\to{3}}}{\left[{2}\cdot{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}={4}$ , find g(3).

Both the functions f and g are continuous at x = 3.

$\displaystyle\lim_{{{x}\to{3}}}{\left[{2}\cdot{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}={4}$

$\displaystyle\Rightarrow{\left[{2}\cdot{f{{\left({3}\right)}}}-{g{{\left({3}\right)}}}\right]}={4}$

$\displaystyle\Rightarrow{2}\cdot{5}-{g{{\left({3}\right)}}}={4}$

$\displaystyle\Rightarrow{10}-{g{{\left({3}\right)}}}={4}$

$\displaystyle\Rightarrow{g{{\left({3}\right)}}}={10}-{4}$

$\displaystyle\Rightarrow{g{{\left({3}\right)}}}={6}$

# Use the definition of continuity and the properties of limits to show that the function $\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{1}}}{{{2}{{x}}^{{2}}-{1}}}$ is continuous at x = 4