Step-by-step explanation:When you substitute the value of 'x' as zero in these expressions, all of them provided a value in the form of [tex]\dfrac{\to0}{\to0}[/tex]We can use the L'Hospital Rule to evaluate these limits. Just find the derivative for both numerator and denominator for these kind of functions.We need to evaluate the following limits:-[tex]1. \ = \lim_{n \to 0}\dfrac{\cos4x-\cos8x}{x^{2}} \\ \\ \\= \dfrac{-\sin4x \times (4)+\sin8x \times 8}{2x} \\ \\ \\= \dfrac{-\cos4x \times (4) \times (4)+\cos8x \times (8)\times (8)}{2} \\ \\ \\= \dfrac{64\cos8x-16\cos4x}{2} \\ \\ \\=\dfrac{64(1)-16(1)}{2} \\ \\ \\=\dfrac{48}{2} = 24[/tex][tex]2. \ \lim_{x \to 0} \dfrac{1+x^{2}-\cos4x}{x \sin{x}} \\ \\ \\= \dfrac{0+2x+4sin4x}{\sin{x}+x\cos{x}} \\ \\ \\= \dfrac{2+16\cos{x}}{cos{x}+\cos{x}-x\sin{x}} \\ \\ \\=\dfrac{2+16(1)}{1+1-0} \\ \\ \\=\dfrac{18}{2} = 9[/tex][tex]3. \ \lim_{x \to 0} \dfrac{x\tan{x}}{1-\cos{x}} \\ \\ \\=\dfrac{x\sec^{2}x+\tan{x}}{\sin{x}} \\ \\ \\=\dfrac{\sec^{2}x+2\sec^{2}{x}\tan{x}+\sec^{2}{x}}{-\cos{x}} \\ \\ \\= \dfrac{(1)+2(1)(0)+(1)}{-1} \\ \\ \\= -2[/tex][tex]4. \ \lim_{x \to 0} \dfrac{6^x-3^x-2^x+1}{x} \\ \\ \\= \dfrac{ln(6)6^x-ln(3)3^x-ln(2)2^x}{1} \\ \\ \\= ln(6)-[ln(3)+ln(2)] \\ \\ = ln(6) - ln(6) \\ \\=0[/tex][tex]5. \ \lim_{x \to 0} \dfrac{5^x+3^x-2^x-1}{x} \\ \\ \\= \dfrac{ln(5)5^x+ln(3)3^x-ln(2)2^x}{1} \\ \\ \\= ln(5) + ln(3)-ln(2) \\ \\ = \ln{\left(\dfrac{15}{2}\right)}[/tex][tex]6. \ \lim_{x \to 0} \dfrac{x\sin{x}}{e^{x}+e^{-x}-2} \\ \\ \\= \dfrac{x\cos{x}+\sin{x}}{e^{x}-e^{-x}} \\ \\ \\= \dfrac{\cos{x}-x\sin{x}+\cos{x}}{e^{x}+e^{-x}} \\ \\ \\=\dfrac{1 - 0 +1}{1+1} \\ \\ \\= 1[/tex][tex]7. \ \lim_{x \to 0} \dfrac{(e^{2x}-1)(1-\cos4x)}{\tan2x} \\ \\ \\= \lim_{x \to 0} \dfrac{e^{2x}-e^{2x}\cos{4x}-1+\cos{4x}}{\tan2x} \\ \\ \\= \dfrac{2e^{2x}+4e^{2x}\sin{4x}-2e^{2x}\cos{4x}-4\sin{4x}}{2\sec^{2}{2x}} \\ \\ \\=\dfrac{2+0-2-0}{2(1)} \\ \\ =0[/tex][tex]8. \ \lim_{x \to 0} \dfrac{5^{2+x}-25}{x} \\ \\ \\= \dfrac{25\ln(5)5^{x}}{1} \\ \\ \\=25 \ln{5}[/tex][tex]9. \ \lim_{h \to 0} \dfrac{\cos{(x+h)}-\cos{x}}{h} \\ \\ \\=\dfrac{-sin{(x+h)(0+1)-0}}{1} \\ \\= -\sin{x}[/tex]
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Step-by-step explanation:
When you substitute the value of 'x' as zero in these expressions, all of them provided a value in the form of [tex]\dfrac{\to0}{\to0}[/tex]
We can use the L'Hospital Rule to evaluate these limits. Just find the derivative for both numerator and denominator for these kind of functions.
We need to evaluate the following limits:-
[tex]1. \ = \lim_{n \to 0}\dfrac{\cos4x-\cos8x}{x^{2}} \\ \\ \\= \dfrac{-\sin4x \times (4)+\sin8x \times 8}{2x} \\ \\ \\= \dfrac{-\cos4x \times (4) \times (4)+\cos8x \times (8)\times (8)}{2} \\ \\ \\= \dfrac{64\cos8x-16\cos4x}{2} \\ \\ \\=\dfrac{64(1)-16(1)}{2} \\ \\ \\=\dfrac{48}{2} = 24[/tex]
[tex]2. \ \lim_{x \to 0} \dfrac{1+x^{2}-\cos4x}{x \sin{x}} \\ \\ \\= \dfrac{0+2x+4sin4x}{\sin{x}+x\cos{x}} \\ \\ \\= \dfrac{2+16\cos{x}}{cos{x}+\cos{x}-x\sin{x}} \\ \\ \\=\dfrac{2+16(1)}{1+1-0} \\ \\ \\=\dfrac{18}{2} = 9[/tex]
[tex]3. \ \lim_{x \to 0} \dfrac{x\tan{x}}{1-\cos{x}} \\ \\ \\=\dfrac{x\sec^{2}x+\tan{x}}{\sin{x}} \\ \\ \\=\dfrac{\sec^{2}x+2\sec^{2}{x}\tan{x}+\sec^{2}{x}}{-\cos{x}} \\ \\ \\= \dfrac{(1)+2(1)(0)+(1)}{-1} \\ \\ \\= -2[/tex]
[tex]4. \ \lim_{x \to 0} \dfrac{6^x-3^x-2^x+1}{x} \\ \\ \\= \dfrac{ln(6)6^x-ln(3)3^x-ln(2)2^x}{1} \\ \\ \\= ln(6)-[ln(3)+ln(2)] \\ \\ = ln(6) - ln(6) \\ \\=0[/tex]
[tex]5. \ \lim_{x \to 0} \dfrac{5^x+3^x-2^x-1}{x} \\ \\ \\= \dfrac{ln(5)5^x+ln(3)3^x-ln(2)2^x}{1} \\ \\ \\= ln(5) + ln(3)-ln(2) \\ \\ = \ln{\left(\dfrac{15}{2}\right)}[/tex]
[tex]6. \ \lim_{x \to 0} \dfrac{x\sin{x}}{e^{x}+e^{-x}-2} \\ \\ \\= \dfrac{x\cos{x}+\sin{x}}{e^{x}-e^{-x}} \\ \\ \\= \dfrac{\cos{x}-x\sin{x}+\cos{x}}{e^{x}+e^{-x}} \\ \\ \\=\dfrac{1 - 0 +1}{1+1} \\ \\ \\= 1[/tex]
[tex]7. \ \lim_{x \to 0} \dfrac{(e^{2x}-1)(1-\cos4x)}{\tan2x} \\ \\ \\= \lim_{x \to 0} \dfrac{e^{2x}-e^{2x}\cos{4x}-1+\cos{4x}}{\tan2x} \\ \\ \\= \dfrac{2e^{2x}+4e^{2x}\sin{4x}-2e^{2x}\cos{4x}-4\sin{4x}}{2\sec^{2}{2x}} \\ \\ \\=\dfrac{2+0-2-0}{2(1)} \\ \\ =0[/tex]
[tex]8. \ \lim_{x \to 0} \dfrac{5^{2+x}-25}{x} \\ \\ \\= \dfrac{25\ln(5)5^{x}}{1} \\ \\ \\=25 \ln{5}[/tex]
[tex]9. \ \lim_{h \to 0} \dfrac{\cos{(x+h)}-\cos{x}}{h} \\ \\ \\=\dfrac{-sin{(x+h)(0+1)-0}}{1} \\ \\= -\sin{x}[/tex]