I need help. Anyone can help?

[moonfei]

QQ图片20210512081937.png (38.22 KiB) Viewed 1297 times

[Guest]

For 13, for x negative and going to 0, 1/x goes to "negative infinity" so what is e to a really large negative value? If x is positive, that's a very large number.'

For 14, [tex]x^3+ x^2= x^2(x+ 1)[/tex] so [tex]\sqrt{x^3+ x^2}= \sqrt{x^2}\sqrt{x+ 1}= |x|\sqrt{x+ 1}[/tex].

If x> 0 [tex]\frac{x^2+ x}{\sqrt{x^3+ 1}}= \frac{x^2+ x}{x\sqrt{x+ 1}}= \frac{x+ 1}{\sqrt{x+ 1}}[/tex].

if x< 0 [tex]\frac{x^2+ x}{\sqrt{x^3+ 1}}= \frac{x^2+ x}{-x\sqrt{x+ 1}}= -\frac{x+ 1}{\sqrt{x+ 1}}[/tex]

[Guest]

The problem says "Using the graph". So where is the graph?

[Guest]

[tex]\lim_{x \to 0 ^{- } } f_{1 }(x) =\lim_{x \to 0 ^{- } } \frac{1}{1+ e^{ \frac{1}{x} } }= \frac{ \lim_{x \to 0^{- }}1 }{ \lim_{x \to 0^{- }}(1+ e^{ \frac{1}{x} })} = \frac{ \lim_{x \to 0^{- }}1 }{ \lim_{x \to 0^{- }}1+ \lim_{x \to 0^{- }}e^{ \frac{1}{x} }}[/tex]

[tex]\lim_{x \to 0^{- }}1=1[/tex]

[tex]\lim_{x \to 0^{- }}1=1[/tex]

[tex]\lim_{x \to 0^{- }}e^{ \frac{1}{x} }= 0[/tex]

[tex]\lim_{x \to 0 ^{- } } f_{1 }(x) = 1[/tex]

[tex]\lim_{x \to 0 ^{+ } } f_{1 }(x) =\lim_{x \to 0 ^{+ } } \frac{1}{1+ e^{ \frac{1}{x} } }= \frac{ \lim_{x \to 0^{+ }}1 }{ \lim_{x \to 0^{+ }}(1+ e^{ \frac{1}{x} })} = \frac{ \lim_{x \to 0^{+ }}1 }{ \lim_{x \to 0^{+ }}1+ \lim_{x \to 0^{+ }}e^{ \frac{1}{x} }}[/tex]

[tex]\lim_{x \to 0^{+ }}1=1[/tex]

[tex]\lim_{x \to 0^{+ }}1=1[/tex]

[tex]\lim_{x \to 0^{+ }}e^{ \frac{1}{x} }= \infty[/tex]

[tex]\lim_{x \to 0 ^{+ } } f_{1 }(x) = 0[/tex]

[tex]1 \ne 0[/tex]

[tex]\lim_{x \to 0 ^{- } } f_{1 }(x) \ne \lim_{x \to 0 ^{+ } } f_{1 }(x)[/tex]

[tex]\lim_{x \to 0 } f_{2 }(x) = \lim_{x \to 0 } \frac{ x^{2 }+x }{ \sqrt{ x^{3 } + x^{2 }} } = \lim_{x \to 0 } \frac{ x (x+1) } { \sqrt{ x^{2 }(x+1) }} = \lim_{x \to 0 }\frac{ x (x+1) } { \sqrt{ x^{2 }(x+1) }} = \lim_{x \to 0 } \frac { x (x+1) } { \sqrt{ x^{2 }} \sqrt {x+1} } = \lim_{x \to 0 } \frac { x (x+1) } { |x| \sqrt {x+1} }=\lim_{x \to 0 } \frac { x } { |x| } \cdot \lim_{x \to 0 } \frac { (x+1) } { \sqrt {x+1} }= \frac { 1 } { \sqrt {1} } \cdot \lim_{x \to 0 } \frac { x } { |x| }[/tex]

[tex]\lim_{x \to 0 ^{-} } \frac { x } { |x| } = \lim_{x \to 0 ^{-} } \frac { x } { -x } = \lim_{x \to 0 ^{-} } (-1)=-1[/tex]

[tex]\lim_{x \to 0 ^{-}} f_{2 }(x) =\frac { 1 } { \sqrt {1} } \cdot (-1) = - \frac { 1 } { \sqrt {1} }[/tex]

[tex]\lim_{x \to 0 ^{+} } \frac { x } { |x| } = \lim_{x \to 0 ^{+} } \frac { x } {x } = \lim_{x \to 0 ^{+} } 1=1[/tex]

[tex]\lim_{x \to 0 ^{+}} f_{2 }(x) =\frac { 1 } { \sqrt {1} } \cdot 1 = \frac { 1 } { \sqrt {1} }[/tex]

[tex]- \frac { 1 } { \sqrt {1} } \ne \frac { 1 } { \sqrt {1} }[/tex]

[tex]\lim_{x \to 0 ^{- } } f_{2 }(x) \ne \lim_{x \to 0 ^{+ } } f_{2 }(x)[/tex]