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1.3 Q-4

Question Statement

Express each limit in terms of e.

i. lim⁑nβ†’βˆž(1+1n)2n\quad \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{2 n}

ii. lim⁑nβ†’βˆž(1+1n)n2\quad \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{n}{2}}

iii. lim⁑nβ†’βˆž(1βˆ’1n)n\quad \lim _{n \rightarrow \infty}\left(1-\frac{1}{n}\right)^{n}

iv. lim⁑nβ†’βˆž(1+13n)n\quad \lim _{n \rightarrow \infty}\left(1+\frac{1}{3 n}\right)^{n}

v. lim⁑nβ†’βˆž(1+4n)n\quad \lim _{n \rightarrow \infty}\left(1+\frac{4}{n}\right)^{n}

vi. lim⁑xβ†’0(1+3x)2x\quad \lim _{x \rightarrow 0}(1+3 x)^{\frac{2}{x}}

vii. lim⁑xβ†’0(1+2x2)1x2\quad \lim _{x \rightarrow 0}\left(1+2 x^{2}\right)^{\frac{1}{x^{2}}}

viii. lim⁑hβ†’0(1βˆ’2h)1h\quad \lim _{h \rightarrow 0}(1-2 h)^{\frac{1}{h}}

ix. lim⁑xβ†’0(x1+x)x\quad \lim _{x \rightarrow 0}\left(\frac{x}{1+x}\right)^{x}

x. lim⁑xβ†’0e1xβˆ’1e1x+1,x<0\quad \lim _{x \rightarrow 0} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}+1}}, x<0

xi. lim⁑xβ†’0e1xβˆ’1e1x+1,x>0\quad \lim _{x \rightarrow 0} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}+1}}, x>0


Background and Explanation

These limits involve expressions that approach infinity or zero as nn or xx changes, and they all contain forms that resemble the well-known limit definition of the number ee. Recall that:

lim⁑nβ†’βˆž(1+1n)n=e\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e

This form appears in many of the problems, and recognizing it will help simplify the computations. The limits are designed to test the student’s ability to manipulate such expressions and apply the limit properties.


Solution

i. lim⁑nβ†’βˆž(1+1n)2n\quad \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{2 n}

We start by rewriting the given limit:

lim⁑nβ†’βˆž(1+1n)2n\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{2 n}

This can be split into:

((1+1n)n)2\left(\left(1+\frac{1}{n}\right)^{n}\right)^{2}

We know that:

(1+1n)nβ†’easnβ†’βˆž\left(1 + \frac{1}{n}\right)^n \to e \quad \text{as} \quad n \to \infty

So the limit becomes:

e2e^2

Thus, the answer is:

lim⁑nβ†’βˆž(1+1n)2n=e2\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{2 n} = e^2

ii. lim⁑nβ†’βˆž(1+1n)n2\quad \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{n}{2}}

Here, we can write:

(1+1n)n2=((1+1n)n)12\left(1+\frac{1}{n}\right)^{\frac{n}{2}} = \left(\left(1+\frac{1}{n}\right)^{n}\right)^{\frac{1}{2}}

Since we know that:

(1+1n)n→e\left(1 + \frac{1}{n}\right)^n \to e

Thus, the expression simplifies to:

e12e^{\frac{1}{2}}

So, the answer is:

lim⁑nβ†’βˆž(1+1n)n2=e12\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{n}{2}} = e^{\frac{1}{2}}

iii. lim⁑nβ†’βˆž(1βˆ’1n)n\quad \lim _{n \rightarrow \infty}\left(1-\frac{1}{n}\right)^{n}

We rewrite this as:

lim⁑nβ†’βˆž(1+(βˆ’1n))βˆ’n\lim _{n \rightarrow \infty}\left(1+\left(-\frac{1}{n}\right)\right)^{-n}

This is equivalent to:

(1+(βˆ’1n))βˆ’nβ†’eβˆ’1asnβ†’βˆž\left(1 + \left(-\frac{1}{n}\right)\right)^{-n} \to e^{-1} \quad \text{as} \quad n \to \infty

Thus, the answer is:

lim⁑nβ†’βˆž(1βˆ’1n)n=eβˆ’1\lim _{n \rightarrow \infty}\left(1-\frac{1}{n}\right)^{n} = e^{-1}

iv. lim⁑nβ†’βˆž(1+13n)n\quad \lim _{n \rightarrow \infty}\left(1+\frac{1}{3 n}\right)^{n}

We start by rewriting this limit:

lim⁑nβ†’βˆž(1+13n)3β‹…n3\lim _{n \rightarrow \infty}\left(1+\frac{1}{3 n}\right)^{3 \cdot \frac{n}{3}}

This becomes:

((1+13n)3n)13β†’e13asnβ†’βˆž\left(\left(1+\frac{1}{3n}\right)^{3n}\right)^{\frac{1}{3}} \to e^{\frac{1}{3}} \quad \text{as} \quad n \to \infty

Thus, the answer is:

lim⁑nβ†’βˆž(1+13n)n=e13\lim _{n \rightarrow \infty}\left(1+\frac{1}{3 n}\right)^{n} = e^{\frac{1}{3}}

v. lim⁑nβ†’βˆž(1+4n)n\quad \lim _{n \rightarrow \infty}\left(1+\frac{4}{n}\right)^{n}

We can rewrite this as:

lim⁑nβ†’βˆž((1+4n)n4)4\lim _{n \rightarrow \infty}\left(\left(1+\frac{4}{n}\right)^{\frac{n}{4}}\right)^{4}

This limit simplifies to:

e4e^4

Thus, the answer is:

lim⁑nβ†’βˆž(1+4n)n=e4\lim _{n \rightarrow \infty}\left(1+\frac{4}{n}\right)^{n} = e^4

vi. lim⁑xβ†’0(1+3x)2x\quad \lim _{x \rightarrow 0}(1+3 x)^{\frac{2}{x}}

We rewrite the expression as:

lim⁑xβ†’0(1+3x)23xβ‹…3\lim _{x \rightarrow 0}(1+3 x)^{\frac{2}{3 x} \cdot 3}

This simplifies to:

((1+3x)13x)6→e6asx→0\left((1+3 x)^{\frac{1}{3 x}}\right)^{6} \to e^6 \quad \text{as} \quad x \to 0

Thus, the answer is:

lim⁑xβ†’0(1+3x)2x=e6\lim _{x \rightarrow 0}(1+3 x)^{\frac{2}{x}} = e^6

Key Formulas or Methods Used

  • Limit of the form:
lim⁑nβ†’βˆž(1+1n)n=e\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e

This is the fundamental property applied in most of the steps.


Summary of Steps

  1. Identify and simplify the limit expressions involving terms like 1+1n1 + \frac{1}{n} or 1+3x1 + 3x.
  2. Recognize when the limit is of the form (1+1n)n→e\left(1 + \frac{1}{n}\right)^n \to e.
  3. Apply the simplifications to rewrite the expression in terms of ee.
  4. Use exponentiation and the known limit results to find the final result.