1] limits- lim ( e^tan h - 1)/h
h->0
2] lim (e^(x+h)² - e^x²)/h
h->0
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We use the definition of e^x to solve this problem.
![e^x = \lim_{n \to \infty} (1+\frac{x}{n})^n\\\\= 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!}+....\infty\\\\e^x=1 + x(1+\frac{x}{2!}+\frac{x^2}{3!}+....+\frac{x^{n-1}}{n!}+....\infty)\\\\1)\\ \lim_{h \to 0}\ \frac{e^{tan\ h}-1}{h} = \lim_{h \to 0}\ \frac{1}{h}[ 1+tan\ h(1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty) - 1]\\\\=\lim_{h \to 0}\ \frac{tan\ h}{h}[ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty]\\\\=1*\lim_{h \to 0}\ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty\\\\=1\\](https://tex.z-dn.net/?f=e%5Ex+%3D++%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%281%2B%5Cfrac%7Bx%7D%7Bn%7D%29%5En%5C%5C%5C%5C%3D+1%2Bx%2B%5Cfrac%7Bx%5E2%7D%7B2%21%7D%2B%5Cfrac%7Bx%5E3%7D%7B3%21%7D%2B....%2B%5Cfrac%7Bx%5En%7D%7Bn%21%7D%2B....%5Cinfty%5C%5C%5C%5Ce%5Ex%3D1+%2B+x%281%2B%5Cfrac%7Bx%7D%7B2%21%7D%2B%5Cfrac%7Bx%5E2%7D%7B3%21%7D%2B....%2B%5Cfrac%7Bx%5E%7Bn-1%7D%7D%7Bn%21%7D%2B....%5Cinfty%29%5C%5C%5C%5C1%29%5C%5C+%5Clim_%7Bh+%5Cto+0%7D%5C+%5Cfrac%7Be%5E%7Btan%5C+h%7D-1%7D%7Bh%7D+%3D+%5Clim_%7Bh+%5Cto+0%7D%5C+%5Cfrac%7B1%7D%7Bh%7D%5B+1%2Btan%5C+h%281%2B%5Cfrac%7Btan%5C+h%7D%7B2%21%7D%2B%5Cfrac%7Btan%5E2%5C+h%7D%7B3%21%7D%2B....%5Cinfty%29+-+1%5D%5C%5C%5C%5C%3D%5Clim_%7Bh+%5Cto+0%7D%5C+%5Cfrac%7Btan%5C+h%7D%7Bh%7D%5B+1%2B%5Cfrac%7Btan%5C+h%7D%7B2%21%7D%2B%5Cfrac%7Btan%5E2%5C+h%7D%7B3%21%7D%2B....%5Cinfty%5D%5C%5C%5C%5C%3D1%2A%5Clim_%7Bh+%5Cto+0%7D%5C+1%2B%5Cfrac%7Btan%5C+h%7D%7B2%21%7D%2B%5Cfrac%7Btan%5E2%5C+h%7D%7B3%21%7D%2B....%5Cinfty%5C%5C%5C%5C%3D1%5C%5C "e^x = \lim_{n \to \infty} (1+\frac{x}{n})^n\\\\= 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!}+....\infty\\\\e^x=1 + x(1+\frac{x}{2!}+\frac{x^2}{3!}+....+\frac{x^{n-1}}{n!}+....\infty)\\\\1)\\ \lim_{h \to 0}\ \frac{e^{tan\ h}-1}{h} = \lim_{h \to 0}\ \frac{1}{h}[ 1+tan\ h(1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty) - 1]\\\\=\lim_{h \to 0}\ \frac{tan\ h}{h}[ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty]\\\\=1*\lim_{h \to 0}\ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty\\\\=1\\")
we use the following limits.
===========================
![\lim_{h \to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}\\\\= \lim_{h \to 0} \frac{1}{h}*[e^{x^2+2xh+h^2}-e^{x^2}]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[e^{2xh+h^2}-1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[1+2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty\ \ -1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty]\\\\=e^{x^2}*\lim_{h \to 0} [2x+h+\frac{h(2x+h)^2}{2!}+\frac{h^2(2x+h)^3}{3!}+....\infty]\\\\=e^{x^2}*2x\\](https://tex.z-dn.net/?f=%5Clim_%7Bh+%5Cto+0%7D+%5Cfrac%7Be%5E%7B%28x%2Bh%29%5E2%7D-e%5E%7Bx%5E2%7D%7D%7Bh%7D%5C%5C%5C%5C%3D+%5Clim_%7Bh+%5Cto+0%7D+%5Cfrac%7B1%7D%7Bh%7D%2A%5Be%5E%7Bx%5E2%2B2xh%2Bh%5E2%7D-e%5E%7Bx%5E2%7D%5D%5C%5C%5C%5C%3De%5E%7Bx%5E2%7D%2A%5Clim_%7Bh+%5Cto+0%7D+%5Cfrac%7B1%7D%7Bh%7D%2A%5Be%5E%7B2xh%2Bh%5E2%7D-1%5D%5C%5C%5C%5C%3De%5E%7Bx%5E2%7D%2A%5Clim_%7Bh+%5Cto+0%7D+%5Cfrac%7B1%7D%7Bh%7D%2A%5B1%2B2xh%2Bh%5E2%2B%5Cfrac%7B%282xh%2Bh%5E2%29%5E2%7D%7B2%21%7D%2B%5Cfrac%7B%282xh%2Bh%5E2%29%5E3%7D%7B3%21%7D%2B....%5Cinfty%5C+%5C+-1%5D%5C%5C%5C%5C%3De%5E%7Bx%5E2%7D%2A%5Clim_%7Bh+%5Cto+0%7D+%5Cfrac%7B1%7D%7Bh%7D%2A%5B2xh%2Bh%5E2%2B%5Cfrac%7B%282xh%2Bh%5E2%29%5E2%7D%7B2%21%7D%2B%5Cfrac%7B%282xh%2Bh%5E2%29%5E3%7D%7B3%21%7D%2B....%5Cinfty%5D%5C%5C%5C%5C%3De%5E%7Bx%5E2%7D%2A%5Clim_%7Bh+%5Cto+0%7D+%5B2x%2Bh%2B%5Cfrac%7Bh%282x%2Bh%29%5E2%7D%7B2%21%7D%2B%5Cfrac%7Bh%5E2%282x%2Bh%29%5E3%7D%7B3%21%7D%2B....%5Cinfty%5D%5C%5C%5C%5C%3De%5E%7Bx%5E2%7D%2A2x%5C%5C "\lim_{h \to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}\\\\= \lim_{h \to 0} \frac{1}{h}*[e^{x^2+2xh+h^2}-e^{x^2}]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[e^{2xh+h^2}-1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[1+2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty\ \ -1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty]\\\\=e^{x^2}*\lim_{h \to 0} [2x+h+\frac{h(2x+h)^2}{2!}+\frac{h^2(2x+h)^3}{3!}+....\infty]\\\\=e^{x^2}*2x\\")
the expression on the RHS simplifies as all the terms except tend to zero, as they contain h.
we use the following limits.
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the expression on the RHS simplifies as all the terms except tend to zero, as they contain h.
Arunav5:
sir is there any other easier meathod its a 1 mark question.
take limit of h ->0 for derivative of numerator divided by derivative of denominator.
1) lim h->0 e^tan h * sec^2 h / 1 => e^0 * sec^2 0 = 1
Another way: given limit is the definition of e^tanh at h = 0. So evaluate its derivative and substitute h = 0.
2) same as above two methods.. given problem is same as the definition of the derivative of e^{x^2}. wrt x. so find it directly.
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