1) 2) 3) 4) 5) 6) 7) 8) 9)

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1) \lim_{x \to \infty} \frac{100}{x}

2) \lim_{x \to \infty} (5-\frac{40}{3x^{2}} )

3) \lim_{x \to \infty} \frac{4x-10}{3+x}

4)\lim_{x \to \infty} (\frac{5x^{3}-7x}{x^{2}+8x})

5) \lim_{x \to 1} \frac{x^{2}-1}{x-1}

6) \lim_{x \to 0} \frac{sin 4x}{2x}

7)\lim_{x \to 0} \frac{tg 5x}{sin15x}

8) \lim_{x \to 0} (1+3x)^{\frac{1}{2}}

9) \lim_{x \to \infty} (1-\frac{1}{2x})^{x}

Алгебра | 58 просмотров
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Правильный ответ
image\infty} \frac{100}{x}=0" alt="lim_{x->\infty} \frac{100}{x}=0" align="absmiddle" class="latex-formula">

image\infty} (5-\frac{40}{3x^2})=5-0=5" alt="lim_{x->\infty} (5-\frac{40}{3x^2})=5-0=5" align="absmiddle" class="latex-formula">

image\infty} \frac{4x-10}{3+x}=lim_{x->\infty} \frac{4-\frac{10}{x}}{\frac{3}{x}+1}=\\\\ \frac{4-0}{0+1}=4" alt="lim_{x->\infty} \frac{4x-10}{3+x}=lim_{x->\infty} \frac{4-\frac{10}{x}}{\frac{3}{x}+1}=\\\\ \frac{4-0}{0+1}=4" align="absmiddle" class="latex-formula">

image\infty} \frac{5x^3-7x}{x^2+8x}=\\\\lim_{x->\infty} \frac{5x-\frac{7}{x^2}}{1+\frac{8}{x}}=\infty" alt="lim_{x->\infty} \frac{5x^3-7x}{x^2+8x}=\\\\lim_{x->\infty} \frac{5x-\frac{7}{x^2}}{1+\frac{8}{x}}=\infty" align="absmiddle" class="latex-formula">

image1} \frac{x^2-1}{x-1}=lim_{x->1} \frac{(x-1)(x+1)}{x-1}=\\\\lim_{x->1} (x+1)=1+1=2" alt="lim_{x->1} \frac{x^2-1}{x-1}=lim_{x->1} \frac{(x-1)(x+1)}{x-1}=\\\\lim_{x->1} (x+1)=1+1=2" align="absmiddle" class="latex-formula">

image0} \frac{sin (4x)}{2x}=lim_{x->0} (\frac{sin(4x)}{(4x)}*2)=1*2=2" alt="lim_{x->0} \frac{sin (4x)}{2x}=lim_{x->0} (\frac{sin(4x)}{(4x)}*2)=1*2=2" align="absmiddle" class="latex-formula">

image0} \frac{tg(5x)}{sin(15x)}=\\\\lim_{x->0} (\frac{tg(5x)}{5x}*\frac{15x}{sin(15x)}*\frac{1}{3}=\\\\1*1*\frac{1}{3}=\frac{1}{3}" alt="lim_{x->0} \frac{tg(5x)}{sin(15x)}=\\\\lim_{x->0} (\frac{tg(5x)}{5x}*\frac{15x}{sin(15x)}*\frac{1}{3}=\\\\1*1*\frac{1}{3}=\frac{1}{3}" align="absmiddle" class="latex-formula">

image0} (1+3x)^{\frac{1}{2}}=(1+3*0)^{\frac{1}{2}}=1" alt="lim_{x->0} (1+3x)^{\frac{1}{2}}=(1+3*0)^{\frac{1}{2}}=1" align="absmiddle" class="latex-formula">

image\infty}(1-\frac{1}{2x})^x=(lim_{x->\infty} (1-\frac{1}{2x})^{-2x})^{-\frac{1}{2}}=\\\\e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}" alt="lim_{x->\infty}(1-\frac{1}{2x})^x=(lim_{x->\infty} (1-\frac{1}{2x})^{-2x})^{-\frac{1}{2}}=\\\\e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}" align="absmiddle" class="latex-formula">
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