We read your photo as:

Evaluate the following limits: 1. lim x->1 (x-1)/(x-1) 2. lim x->1 (x2-1)/(x2-1) 3. lim x->3 (x3-27)/(x-9) 4. lim x->0 ((3+x)2-9)/x

Mathematics

ScanSolve AI Answer

Step-by-step solution

31 steps

Here are the solutions to the problems.

1.

  1. Substitute x=1x=1 into the expression.

    1−11−1=00\frac{1-1}{1-1} = \frac{0}{0}

    This is an indeterminate form.

  2. Simplify the expression.

    For x≠1x \neq 1, the expression x−1x−1\frac{x-1}{x-1} simplifies to 1.

  3. Evaluate the limit.

    lim⁡x→1x−1x−1=lim⁡x→11=1\lim_{x \to 1} \frac{x-1}{x-1} = \lim_{x \to 1} 1 = 1

    The limit is 1\boxed{1}.

    2.

  4. Substitute x=1x=1 into the expression.

    12−112−1=1−11−1=00\frac{1^2-1}{1^2-1} = \frac{1-1}{1-1} = \frac{0}{0}

    This is an indeterminate form.

  5. Simplify the expression.

    For x≠1x \neq 1, the expression x2−1x2−1\frac{x^2-1}{x^2-1} simplifies to 1.

  6. Evaluate the limit.

    lim⁡x→1x2−1x2−1=lim⁡x→11=1\lim_{x \to 1} \frac{x^2-1}{x^2-1} = \lim_{x \to 1} 1 = 1

    The limit is 1\boxed{1}.

    3.

  7. Substitute x=3x=3 into the expression.

    33−273−9=27−27−6=0−6=0\frac{3^3-27}{3-9} = \frac{27-27}{-6} = \frac{0}{-6} = 0

    This is not an indeterminate form.

  8. Evaluate the limit.

    lim⁡x→3x3−27x−9=0\lim_{x \to 3} \frac{x^3-27}{x-9} = 0

    The limit is 0\boxed{0}.

    4.

  9. Substitute x=0x=0 into the expression.

    (3+0)2−90=32−90=9−90=00\frac{(3+0)^2-9}{0} = \frac{3^2-9}{0} = \frac{9-9}{0} = \frac{0}{0}

    This is an indeterminate form.

  10. Expand the numerator.

    (3+x)2−9=(9+6x+x2)−9=6x+x2(3+x)^2 - 9 = (9 + 6x + x^2) - 9 = 6x + x^2
  11. Simplify the expression.

    6x+x2x=x(6+x)x=6+xfor x≠0\frac{6x+x^2}{x} = \frac{x(6+x)}{x} = 6+x \quad \text{for } x \neq 0
  12. Evaluate the limit.

    lim⁡x→0(6+x)=6+0=6\lim_{x \to 0} (6+x) = 6+0 = 6

    The limit is 6\boxed{6}.

    5.

  13. Substitute x=1x=1 into the expression.

    13−112−3(1)+2=1−11−3+2=00\frac{1^3-1}{1^2-3(1)+2} = \frac{1-1}{1-3+2} = \frac{0}{0}

    This is an indeterminate form.

  14. Factor the numerator and denominator.

    The numerator is a difference of cubes: x3−1=(x−1)(x2+x+1)x^3-1 = (x-1)(x^2+x+1).
    The denominator is a quadratic: x2−3x+2=(x−1)(x−2)x^2-3x+2 = (x-1)(x-2).

  15. Simplify the expression.

    (x−1)(x2+x+1)(x−1)(x−2)=x2+x+1x−2for x≠1\frac{(x-1)(x^2+x+1)}{(x-1)(x-2)} = \frac{x^2+x+1}{x-2} \quad \text{for } x \neq 1
  16. Evaluate the limit.

    lim⁡x→1x2+x+1x−2=12+1+11−2=3−1=−3\lim_{x \to 1} \frac{x^2+x+1}{x-2} = \frac{1^2+1+1}{1-2} = \frac{3}{-1} = -3

    The limit is −3\boxed{-3}.

    6.

  17. Substitute x=5x=5 into the expression.

    52−7(5)+1052−25=25−35+1025−25=00\frac{5^2-7(5)+10}{5^2-25} = \frac{25-35+10}{25-25} = \frac{0}{0}

    This is an indeterminate form.

  18. Factor the numerator and denominator.

    The numerator is a quadratic: x2−7x+10=(x−2)(x−5)x^2-7x+10 = (x-2)(x-5).
    The denominator is a difference of squares: x2−25=(x−5)(x+5)x^2-25 = (x-5)(x+5).

  19. Simplify the expression.

    (x−2)(x−5)(x−5)(x+5)=x−2x+5for x≠5\frac{(x-2)(x-5)}{(x-5)(x+5)} = \frac{x-2}{x+5} \quad \text{for } x \neq 5
  20. Evaluate the limit.

    lim⁡x→5x−2x+5=5−25+5=310\lim_{x \to 5} \frac{x-2}{x+5} = \frac{5-2}{5+5} = \frac{3}{10}

    The limit is 310\boxed{\frac{3}{10}}.

    For questions 7-11, we will find the derivative dydx\frac{dy}{dx}.

    7.

  21. Rewrite the function.

    y=1cos⁡x=sec⁡xy = \frac{1}{\cos x} = \sec x
  22. Differentiate yy with respect to xx.

    dydx=ddx(sec⁡x)=sec⁡xtan⁡x\frac{dy}{dx} = \frac{d}{dx}(\sec x) = \sec x \tan x

    The derivative is sec⁡xtan⁡x\boxed{\sec x \tan x}.

    8.

  23. Rewrite the function.

    y=1sin⁡x=csc⁡xy = \frac{1}{\sin x} = \csc x
  24. Differentiate yy with respect to xx.

    dydx=ddx(csc⁡x)=−csc⁡xcot⁡x\frac{dy}{dx} = \frac{d}{dx}(\csc x) = -\csc x \cot x

    The derivative is −csc⁡xcot⁡x\boxed{-\csc x \cot x}.

    9.

  25. Apply the chain rule.

    Let u=x2+4u = x^2+4. Then y=u4y = u^4.

    dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
  26. Find dydu\frac{dy}{du} and dudx\frac{du}{dx}.

    dydu=ddu(u4)=4u3\frac{dy}{du} = \frac{d}{du}(u^4) = 4u^3 dudx=ddx(x2+4)=2x\frac{du}{dx} = \frac{d}{dx}(x^2+4) = 2x
  27. Substitute back u=x2+4u = x^2+4.

    dydx=4(x2+4)3⋅(2x)=8x(x2+4)3\frac{dy}{dx} = 4(x^2+4)^3 \cdot (2x) = 8x(x^2+4)^3

    The derivative is 8x(x2+4)3\boxed{8x(x^2+4)^3}.

    10.

  28. Differentiate yy with respect to xx.

    dydx=ddx(sin⁡x)=cos⁡x\frac{dy}{dx} = \frac{d}{dx}(\sin x) = \cos x

    The derivative is cos⁡x\boxed{\cos x}.

    11.

  29. Apply the chain rule.

    Let u=x2+4u = x^2+4. Then y=u3y = u^3.

    dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
  30. Find dydu\frac{dy}{du} and dudx\frac{du}{dx}.

    dydu=ddu(u3)=3u2\frac{dy}{du} = \frac{d}{du}(u^3) = 3u^2 dudx=ddx(x2+4)=2x\frac{du}{dx} = \frac{d}{dx}(x^2+4) = 2x
  31. Substitute back u=x2+4u = x^2+4.

    dydx=3(x2+4)2⋅(2x)=6x(x2+4)2\frac{dy}{dx} = 3(x^2+4)^2 \cdot (2x) = 6x(x^2+4)^2

    The derivative is 6x(x2+4)2\boxed{6x(x^2+4)^2}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp