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Textbook 2 Videos \( [+] \) Evaluate the limit: \( \lim _{h \rightarrow 0} \frac{(1+h)^{2}-1}{h} \) Question Help: Video \( \square \) Message instructor Submit Question Jump to Answer

Ask by Nunez Cruz. in the United States
Jan 27,2025

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Tutor-Verified Answer

Answer

The limit is 2.

Solution

To evaluate the limit: \[ \lim_{h \rightarrow 0} \frac{(1+h)^{2} - 1}{h} \] let's follow the step-by-step process. ### Step 1: Expand the Numerator First, expand \((1 + h)^2\): \[ (1 + h)^2 = 1 + 2h + h^2 \] Now, subtract 1 from the expanded form: \[ (1 + 2h + h^2) - 1 = 2h + h^2 \] So, the expression becomes: \[ \frac{2h + h^2}{h} \] ### Step 2: Simplify the Expression Factor \(h\) out of the numerator: \[ \frac{h(2 + h)}{h} \] Now, cancel out the \(h\) in the numerator and denominator (assuming \(h \neq 0\)): \[ 2 + h \] ### Step 3: Evaluate the Limit Now, take the limit as \(h\) approaches 0: \[ \lim_{h \rightarrow 0} (2 + h) = 2 + 0 = 2 \] ### Conclusion The limit is: \[ 2 \]

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To evaluate the limit \( \lim _{h \rightarrow 0} \frac{(1+h)^{2}-1}{h} \), let's first simplify the expression. Expanding \( (1+h)^{2} \) gives \( 1 + 2h + h^{2} \). Therefore, the expression becomes: \[ \frac{(1 + 2h + h^{2}) - 1}{h} = \frac{2h + h^{2}}{h} = 2 + h \] Now, as \( h \) approaches 0, the limit simplifies to \( 2 + 0 = 2 \). Thus, \[ \lim _{h \rightarrow 0} \frac{(1+h)^{2}-1}{h} = 2. \] This result is a fundamental concept in calculus, representing the derivative of the function \( f(x) = x^{2} \) at the point \( x = 1 \). It's a key building block for understanding rates of change! Understanding limits like this is crucial to gaining a solid grasp of calculus concepts. You can explore them further through hands-on activities or interactive online courses that visualize these ideas, making learning both fun and effective. Consider resources like Khan Academy or MIT OpenCourseWare for comprehensive lessons packed with practical examples and visual aids!

Related Questions

Multiple Choice Identify the choice that best completes the statement or answers the question. Find any points of discontinuity for the rational function. 1. \( y=\frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) a. \( x=-5, x=-2 \) b. \( x=5, x=2 \) c. \( x=-7, x=2, x=-9 \) d. \( x=7, x=-2, x=9 \) 2. \( y=\frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) a. \( x=-5, x=3 \) b. \( x=7, x=4, x=2 \) c. \( x=-7, x=-4, x=-2 \) d. \( x=5, x=-3 \) 3. \( y=\frac{x+4}{x^{2}+8 x+15} \) a. \( x=-5, x=-3 \) b. \( x=-4 \) c. \( x=-5, x=3 \) d. \( x=5, x=3 \) 4. \( y=\frac{x-3}{x^{2}+3 x-10} \) a. \( x=-5, x=2 \) b. \( x=5, x=-2 \) c. \( x=3 \) d. \( x \) \( =-5, x=-2 \) 6. What are the points of discontinuity? Are they all removable? \[ y=\frac{(x-4)}{x^{2}-13 x+36} \] a. \( x=-9, x=-4, x=8 \); yes b. \( x=1, x=8, x= \) -8; no c. \( x=9, x=4 \); no d. \( x=-9, x=-4 \); no 7. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \). a. asymptote: \( x=2 \) and hole: \( x=-5 \) b. asymptotes: \( x=-2 \) and hole: \( x=-5 \) c. asymptote: \( x=-2 \) and hole: \( x=5 \) d. asymptote: \( x=-2 \) and hole: \( x=-2 \) a. \( x=-3, x=-8 \); no b. \( x=5, x=-7, x=1 \); no c. \( x=-5, x=7, x=-1 \); yes d. \( x=3, x=8 \); yes 8. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{x+1}{x^{2}+6 x+5} \). a. asymptote: \( x=-1 \) and hole: \( x=-1 \) b. asymptote: \( x=-1 \) and hole: \( x=-1 \) c. asymptotes: \( x=-1,-1 \) and hole: \( x=-1 \) d. asymptote: -5 and hole: \( x=-1 \), 9. Find the horizontal asymptote of the graph of \( y=\frac{7 x^{6}+7 x+3}{9 x^{5}+7 x+3} \). a. \( y=0 \) b. \( y=\frac{7}{9} \) c. no horizontal asymptote d. \( y=\frac{6}{5} \)
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